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WebAssign General Chemistry Atoms First (Homework)

James Finch

Chemistry - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 24

Due : Tuesday, March 24, 2020 20:00 EDT

Last Saved : n/a Saving...  ()

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Calculate the number of moles of potassium permanganate (KMnO4) corresponding to 213.7 g of the substance.
WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

1.352

 mol




Solution or Explanation

  • Step 1 of 4

    Use the periodic table to determine the average atomic mass of each element in the substance.

    The average atomic mass of potassium (K) = 39.10 amu.

    The average atomic mass of manganese (Mn) = 54.94 amu.

    The average atomic mass of oxygen (O) = 16.00 amu.

  • Step 2 of 4

    Multiply each average atomic mass by the number of atoms representing that element in one formula unit of the substance. Then, sum the resulting values to obtain the formula mass of the substance.

    formula mass of KMnO4 = (# atoms of K)(average atomic mass of K) + (# atoms of Mn)(average atomic mass of Mn) + (# atoms of O)(average atomic mass of O)
     = 
    1 atom of K
    39.10 amu
    atom of K
     + 
    1 atom of Mn
    54.94 amu
    atom of Mn
     + 
    4 atoms of O
    16.00 amu
    atom of O
     = 39.100 amu + 54.940 amu + 64.000 amu
     = 158.04 amu (given to the correct number of significant figures)

  • Step 3 of 4

    Find the molar mass of the substance in grams per mole, which is equal to the formula mass in atomic mass units.

    Molar mass in grams per mole = formula mass in atomic mass units.
    Therefore, the molar mass of KMnO4 is 158.04 g/mol.

  • Step 4 of 4

    Use the molar mass of the substance and dimensional analysis to set up a unit conversion to convert grams to moles, and perform the calculation. (Show the result with the correct number of significant figures.)

    moles of KMnO4 = 213.7 g KMnO4 × 
    1 mol KMnO4
    158.04 g KMnO4
     = 1.352 mol KMnO4
    (Note that the result is expressed with the same number of significant figures as the term with the least number of significant figures.)

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(a) How many moles of CO2 contain 2.91 ✕ 1024 molecules?
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4.83

 mol CO2

(b) What number of moles is equivalent to 6.34 ✕ 1021 atoms of Hg?
WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

0.0105

 mol Hg




Solution or Explanation

  • Step 1 of 4

    (a) How many moles of CO2 contain 2.91 ✕ 1024 molecules?

    Write a conversion factor to change the number of molecules of CO2 to moles of CO2.

    The number of molecules is given. The number of moles must be determined. Therefore, the unit molecules must cancel when the given value is multiplied by the conversion factor.
    1 mol CO2 = 6.02214199 ✕ 1023 molecules CO2
    For this to occur, the number of molecules must be in the denominator (bottom) and the desired unit, moles, must be in the numerator (top).
    1 mol CO2
    6.02214199 ✕ 1023 molecules CO2

  • Step 2 of 4

    Multiply the given number of molecules by the conversion factor.

    Multiply the given number of molecules, 2.91 ✕ 1024, by the conversion factor equivalent to the number of molecules in one mole. To find the answer, divide the number of molecules by 6.02214199 ✕ 1023. The unit molecules will cancel.
    2.91 ✕ 1024 molecules CO2 × 
    1 mol CO2
    6.02214199 ✕ 1023 molecules CO2
     = 4.83 mol CO2

  • Step 3 of 4

    (b) What number of moles is equivalent to 6.34 ✕ 1021 atoms of Hg?

    Write a conversion factor to change the number of atoms of Hg to moles of Hg.

    The number of atoms is given. The number of moles must be determined. Therefore, the unit atoms must cancel when the given value is multiplied by the conversion factor.
    1 mol Hg = 6.02214199 ✕ 1023 atoms Hg
    For this to occur, the number of atoms must be in the denominator (bottom) and the desired unit, moles, must be in the numerator (top).
    1 mol Hg
    6.02214199 ✕ 1023 atoms Hg

  • Step 4 of 4

    Multiply the number of atoms by the conversion factor.

    Multiply the given number of atoms, 6.34 ✕ 1021, by the conversion factor equivalent to the number of atoms in one mole. The unit atoms will cancel.
    6.34 ✕ 1021 atoms Hg × 
    1 mol Hg
    6.02214199 ✕ 1023 atoms Hg
     = 0.0105 mol Hg

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
Write the molecular and empirical formulas for the compound with the following structural formula.
Step 1
Write the given structural formula of the compound.

(No Response)
Marvin JS


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A compound has an empirical formula of CH2O and a molar mass of 180. g/mol. What is the molecular formula?

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C6H12O6





Solution or Explanation

  • Step 1 of 5

    Recall the definition of molar mass.

    molar mass = mass of one mole of a substance (in grams)
    The molar mass of a compound is found by summing the molar masses of its constituent atoms.
    molar mass of a compound in g/mol = formula mass of that compound in amu

  • Step 2 of 5

    Recall the definition of an empirical formula.

    An empirical formula expresses the simplest whole-number ratio of distinct atoms in a compound.

  • Step 3 of 5

    Calculate the molar mass of the given empirical formula CH2O. (Enter unrounded values.)

    formula mass of CH2O = (# C atoms)(atomic mass of C)
     + (# H atoms)(atomic mass of H)
     + (# O atoms)(atomic mass of O)
     = (1)(12.01 amu) + (2)(1.008 amu) + (1)(16.00amu)
     = 30.026 amu
    Therefore, the empirical molar mass of CH2O is 30.026 g/mol.

  • Step 4 of 5

    Calculate the ratio between the given molar mass and the empirical molar mass calculated in Step 3.

    The ratio between the given molar mass of 180. g/mol and the calculated empirical molar mass of 30.026 g/mol is the following.
    180. g/mol
    30.026 g/mol
    		 = 5.995 (Enter an unrounded value. Use at least one more digit than given.)
    ≈6 (Rounded to the nearest whole number.)

  • Step 5 of 5

    Multiply the subscripts in the empirical formula by the calculated ratio from Step 4 to determine the molecular formula.

    [Axiom: multiplication; Skill FMM0250]

    C: 6 × 1 = 6
    H: 6 × 2 = 12
    O: 6 × 1 = 6
    Therefore, the actual molecular formula is:
    C6H12O6

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A forensics laboratory received a sample of white powder that was suspected of being ecstasy (C11H15NO2), heroin (C21H23NO5), morphine (C17H19NO3), or demerol (C15H21NO2). The results of an elemental analysis performed on the substance are 72.84% carbon, 8.56% hydrogen, 5.66% nitrogen, and 12.94% oxygen. What is the identity of the unknown substance?
    





Solution or Explanation

  • Step 1 of 6

    Assume that 100 g of the substance is available. Determine the mass, in grams, of each element in that quantity of substance.

    C: 72.84% of 100 g = 72.84 g C
    H: 8.56% of 100 g = 8.56 g H
    N: 5.66% of 100 g = 5.66 g N
    O: 12.94% of 100 g = 12.94 g O

  • Step 2 of 6

    Use the periodic table to determine the molar mass of each element present in the substance. (Enter an unrounded value.)

    Recall that average atomic mass is reported on the periodic table, and that molar mass in g/mol equals average atomic mass in amu.
    molar mass of C:12.01 g/mol
    molar mass of H:1.008 g/mol
    molar mass of N:14.01 g/mol
    molar mass of O:16.00 g/mol

  • Step 3 of 6

    Calculate the number of moles of each element present in the substance. (Enter an unrounded value. Use at least one more digit than given.)

    C: 72.84 g C×
    1 mol C
    12.01 g C
    		 = 6.0649 mol C
    H: 8.56 g H×
    1 mol H
    1.008 g H
    		 = 8.492 mol H
    N: 5.66 g N×
    1 mol N
    14.01 g N
    		 = 0.404 mol N
    O: 12.94 g O×
    1 mol O
    16.00 g O
    		 = 0.80875 mol O

  • Step 4 of 6

    Divide the number of moles of each element present by the smallest number of moles represented by any of the elements. If the results are very close to positive integers, round them to the nearest integer.

    The smallest number of moles from Step 3 comes from nitrogen (N), 0.4040 mol.
    C: 
    6.0649
     
     
    mol
    0.4040 mol
    		 = 15.0 (rounded to 3 significant figures)
     = 15 (rounded to the nearest integer)
    H: 
    8.492
     
     
    mol
    0.4040 mol
    		 = 21.0 (rounded to 3 significant figures)
     = 21 (rounded to the nearest integer)
    N: 
    0.4040
     
     
    mol
    0.4040 mol
    		 = 1.00 (rounded to 3 significant figures)
     = 1 (rounded to the nearest integer)
    O: 
    0.80875
     
     
    mol
    0.4040 mol
    		 = 2.00 (rounded to 3 significant figures)
     = 2 (rounded to the nearest integer)

  • Step 5 of 6

    Using the relative number of atoms in the formula determined in Step 4, write the empirical formula of the substance.

    The C, H, N, and O atoms in the substance exist in the ratio 15:21:1:2, respectively. The empirical formula is therefore the following.
    C15H21NO2

  • Step 6 of 6

    Compare the empirically determined chemical formula to the possible formulas, and identify the compound.

    The empirically determined chemical formula matches which substance?
    demerol (C15H21NO2)

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A 0.5888 g sample of an unknown acid was titrated with NaOH and found to contain 4.172 ✕ 10−3 mol acid. Another sample of the substance was subjected to elemental analysis, where it was determined that the composition of the unknown material is 67.61% C, 9.90% H, and 22.48% O.
(a) What is the molar mass of the unknown acid?
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141.1

 g/mol

(b) What is the empirical formula of the unknown acid?

chemPad

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C4H7O

(c) What is the molecular formula of the unknown acid?

chemPad

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C8H14O2




Solution or Explanation

  • Step 1 of 10

    (a) What is the molar mass of the unknown acid?

    Calculate the molar mass of the acid.

    Recall that the molar mass of a compound is
    grams X
    moles X
    ;
     therefore, the molar mass of the unknown acid is:
    molar mass = 
    (0.5888 g)
    (4.172 ✕ 10−3 mol)
     = 141.1 g/mol.
    The answer is shown with four significant figures to match the precision of the measured values.

  • Step 2 of 10

    (b) What is the empirical formula of the unknown acid?

    Assume that 100 g of the substance is available. Determine the mass, in grams, of each element in that quantity of substance. (Enter an unrounded value.)

    C: 67.61% of 100 g = 67.61 g C
    H: 9.90% of 100 g = 9.90 g H
    O: 22.48% of 100 g = 22.48 g O

  • Step 3 of 10

    Use the periodic table to determine the average molar mass of each element present in the substance.

    Recall that average atomic mass (in amu) is reported on the periodic table and is equivalent to molar mass (in g/mol).
    average molar mass of C: 12.01 g/mol
    average molar mass of H: 1.008 g/mol
    average molar mass of O: 16.00 g/mol

  • Step 4 of 10

    Calculate the number of moles of each element present in the substance. (Enter an unrounded value. Use at least one more digit than given.)

    C: 67.61 g C × 
    1 mol C
    12.01 g C
     = 5.6295 mol C
    H: 9.90 g H × 
    1 mol H
    1.008 g H
     = 9.821 mol H
    O: 22.48 g O × 
    1 mol O
    16.00 g O
     = 1.405 mol O

  • Step 5 of 10

    Divide the number of moles present in each element by the smallest number of moles represented by any of the elements. If the results are very close to positive integers, round them to the nearest integer.

    The smallest number of moles from Step 4 comes from oxygen (O), 1.4050 mol.
    C: 
    5.6295 mol
     
     
    1.4050 mol
     
     
    		 = 4.007 (rounded to 4 significant figures)
     = 4 (rounded to the nearest integer)
    H: 
    9.821 mol
     
     
    1.4050 mol
     
     
    		 = 6.99 (rounded to 3 significant figures)
     = 7 (rounded to the nearest integer)
    O: 
    1.4050 mol
     
     
    1.4050 mol
     
     
    		 = 1.000 (rounded to 4 significant figures)
     = 1 (rounded to the nearest integer)

  • Step 6 of 10

    Using the relative number of atoms in the formula determined in Step 5, write the empirical formula of the substance.

    The C, H, and O atoms in the substance exist in the ratio 4:7:1, respectively. The empirical formula is therefore the following.
    C4H7O

  • Step 7 of 10

    (c) What is the molecular formula of the unknown acid?

    Using the periodic table and empirical formula of the compound, calculate the molar mass of the empirical formula.

    Empirical formula mass of C4H7O = (4)(atomic mass of C) + (7)(atomic mass of H)
    + (1)(atomic mass of O)
     = (4)(12.01 amu) + (7)(1.008 amu) + (1)(16.00 amu)
     = 71.096 amu
    Formula mass in amu equals molar mass in g/mol, so the molar mass of the empirical formula is 71.096 g/mol.

  • Step 8 of 10

    Divide the empirical molecular mass of the substance by the empirical molecular mass of the fragment, and then round the quotient to the nearest integer.

    141.1 g/mol
    71.096 g/mol
    		 = 1.985 (rounded to 4 significant figures)
     = 2 (rounded to the nearest integer)

  • Step 9 of 10

    Multiply each subscript in the given original empirical formula, C4H7O, by the number calculated in Step 8 to obtain the molecular formula.

    C8H14O2

  • Step 10 of 10

    Summarize the answer.

    (a) The molar mass of the unknown acid is 141.1 g/mol.

    (b) The empirical formula of the unknown acid is the following.
    C4H7O

    (c) The molecular formula of the unknown acid is the following.
    C8H14O2

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Determine whether 3 H2 + N2 → 2 NH3 obeys the law of conservation of mass and justify your response.
Mass (No Response) seenKey

has

 been conserved in this reaction because the total mass of the reactants is (No Response) seenKey

equal to

 the total mass of the products.




Solution or Explanation

  • Step 1 of 5

    Recall the law of conservation of mass.

    When the law is applied to a chemical reaction, it implies that the total mass of the reactants should be equal to the total mass of the products of a chemical reaction. It also means that the total number and type of atoms in the reactants and products should be identical.

  • Step 2 of 5

    Identify the number and types of atoms present in the reactants and products.

    The reactants contain three molecule(s) of H2 and one molecule(s) of N2. The products contain two molecule(s) of NH3. Therefore, the following atoms are present.
    Reactants:
    6 H atom(s) (two from each H2 molecule)
    2 N atom(s) (two from each N2 molecule)

    Products:
    2 N atom(s) (one from each NH3 molecule)
    6 H atom(s) (three from each NH3 molecule)

  • Step 3 of 5

    Find the mass per mole of each reactant and the product. (Enter an unrounded value.)

    formula mass of H2 = (# H atoms)(atomic mass of H)
     = (2)(1.008)
     = 2.016 amu

    molar mass in grams per mole = formula mass in atomic mass units
    Therefore, molar mass of H2 = 2.016 g/mol.

    formula mass of N2 = (# N atoms)(atomic mass of N)
     = (2)(14.01)
     = 28.02 amu

    molar mass in grams per mole = formula mass in atomic mass units
    Therefore, molar mass of N2 = 28.02 g/mol.

    formula mass of NH3 = (# N atoms)(atomic mass of N)+(# H atoms)(atomic mass H)
     = (1)(14.01) + (3)(1.008)
     = 17.034 amu

    molar mass in grams per mole = formula mass in atomic mass units
    Therefore, molar mass of NH3 = 17.034 g/mol.

  • Step 4 of 5

    Using the molar masses calculated in Step 3 and the balanced equation given above, find the mass of each reactant and the product. (Enter an unrounded value.)

    Multiply the molar mass of each reactant and the product by its respective number of moles (coefficient) shown in the balanced equation above.
    3 H2:
    2.016 g
    1 mol
     × 3 mol    		 = 6.048 g
    N2:
    28.02 g
    1 mol
     × 1 mol    		 = 28.02 g
    2 NH3:
    17.034 g
    1 mol
     × 2 mol    		 = 34.068 g

  • Step 5 of 5

    Compare the final mass of the reactants with that of the product and determine if mass was conserved. (Enter an unrounded value.)

    [Definition CRN0024; Axiom: Applying definitions]

    For mass to have been conserved, the combined mass of the two reactants must be equal to the mass of the product.
    mass of 3 H2 + mass N2 = mass 2 NH3
    (6.048 g) + (28.02 g) = 34.068 g
    34.068 g = 34.068 g
    Therefore, mass has been conserved in this reaction.

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Methane gas (CH4) reacts with chlorine gas (Cl2) to produce liquid carbon tetrachloride (CCl4) and hydrogen chloride gas (HCl). The balanced chemical equation for the reaction is the following.
CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)
Calculate and compare the numbers of atoms of each element on the reactant and product sides of this chemical equation.
In the chemical equation given, (No Response) seenKey

1

 C atom(s), (No Response) seenKey

4

 H atom(s), and (No Response) seenKey

8

 Cl atom(s) are present on the reactant side and (No Response) seenKey

1

 C atom(s), (No Response) seenKey

4

 H atom(s), and (No Response) seenKey

8

 Cl atom(s) are present on the product side.




Solution or Explanation

  • Step 1 of 9

    Write the given chemical equation.

    The chemical equation is given as the following. (Include states-of-matter under the given conditions in your answer.)
    CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)

  • Step 2 of 9

    Identify the reactants and the products from the chemical equation. (Separate substances in a list with a comma. Omit states-of-matter from your answer.)

    (a) The chemical formulas of the reactants are the following.
    CH4, Cl2

    (b) The chemical formulas of the products are the following.
    CCl4, HCl

  • Step 3 of 9

    Determine the subscripts and coefficient associated with each reactant. Remember that the absence of a subscript implies a subscript of 1, and the absence of a coefficient implies a coefficient of 1.

    (i)    CH4: The subscript of C is 1, the subscript of H is 4 and the coefficient of CH4 is 1.

    (ii)    4 Cl2: The subscript of Cl is 2, and the coefficient of Cl2 is 4.

  • Step 4 of 9

    Calculate the number of atoms of each element in each reactant by multiplying the subscript of the element symbol in the chemical formula with the coefficient of that chemical formula in the chemical equation.

    (i)    CH4:
    number of C atoms in CH4 = (subscript of C)(coefficient of CH4) = (1)(1) = 1 C atom(s)
    number of H atoms in CH4 = (subscript of H)(coefficient of CH4) = (4)(1) = 4 H atom(s)

    (ii)    4 Cl2:
    number of Cl atoms in 4 Cl2 = (subscript of Cl)(coefficient of Cl2) = (2)(4) = 8 Cl atom(s)

  • Step 5 of 9

    Sum the total number of atoms of each element present on the reactant side of the equation.

    C atoms occur in one reactant. The total number of C atoms is 1.
    H atoms occur in one reactant. The total number of H atoms is 4.
    Cl atoms occur in one reactant. The total number of Cl atoms is 8.

    Therefore, in the chemical equation given, the reactants contain 1 C atom(s), 4 H atom(s), and 8 Cl atom(s).

  • Step 6 of 9

    Determine the subscripts and coefficient associated with each product. Remember that the absence of a subscript implies a subscript of 1, and the absence of a coefficient implies a coefficient of 1.

    (i)    CCl4: The subscript of C is 1, the subscript of Cl is 4, and the coefficient of CCl4 is 1.

    (ii)    4 HCl: The subscript of H is 1, the subscript of Cl is 1, and the coefficient of HCl is 4.

  • Step 7 of 9

    Calculate the number of atoms of each element in each product by multiplying the subscript of the element symbol in the chemical formula with the coefficient of that chemical formula in the chemical equation.

    (i)    CCl4:
    number of C atoms in CCl4 = (subscript of C)(coefficient of CCl4) = (1)(1) = 1 C atom(s)
    number of Cl atoms in CCl4 = (subscript of Cl)(coefficient of CCl4) = (4)(1) = 4 Cl atom(s)

    (ii)    4 HCl:
    number of H atoms in 4 HCl = (subscript of H)(coefficient of HCl) = (1)(4) = 4 H atom(s)
    number of Cl atoms in 4 HCl = (subscript of Cl)(coefficient of HCl) = (1)(4) = 4 Cl atom(s)

  • Step 8 of 9

    Sum the total number of atoms of each element present on the product side of the equation.

    C atoms occur in one product. The total number of C atoms is 1.
    H atoms occur in one product. The total number of H atoms is 4.
    Cl atoms occur in two products. The total number of Cl atoms = 4 atoms in CCl4 + 4 atoms in 4 HCl = 8 Cl atoms.

    Therefore, in the chemical equation given, the products contain 1 C atom(s), 4 H atom(s), and 8 Cl atom(s).

  • Step 9 of 9

    From Steps 5 and 8 above, compare the number of atoms of each element on the reactant side of the chemical equation with the number of atoms of each element on the product side of the chemical equation.

    From Step 5, the reactant side of the equation contains 1 C atom(s), 4 H atom(s), and 8 Cl atom(s).

    From Step 8, the product side of the equation contains 1 C atom(s), 4 H atom(s), and 8 Cl atom(s).

    In this case, the number of atoms of each element is the same on both the reactant and product sides of the given chemical equation.

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Four moles of hydrogen iodide (HI) gas react with one mole of oxygen (O2) gas to produce two moles of iodine gas (I2) and two moles of water gas (H2O). Using this description of the reaction, write the corresponding chemical equation, showing the appropriate coefficients for each reactant and product. (Include states-of-matter under the given conditions in your answer.)

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4 HI(g) + O2(g)  →  2 I2(g) + 2 H2O(g)





Solution or Explanation

  • Step 1 of 6

    Identify the reactants as the starting substances that react with each other.

    The reactants, or the starting substances that react with each other, are hydrogen iodide (HI) gas and oxygen (O2) gas.

  • Step 2 of 6

    Write the reactants in terms of their chemical formulas and physical states. Group the reactants together with an addition (+) sign. (Write the reactants in the order in which they are given.)

    In terms of their chemical formulas and physical states, the reactants can be written as the following.
    HI(g) + O2(g)

  • Step 3 of 6

    Identify the products as the substances that are produced as a result of the reaction.

    The products, or the substances produced, are iodine gas (I2) and water gas (H2O).

  • Step 4 of 6

    Write the products in terms of their chemical formulas and physical states. Group the products together with an addition (+) sign. (Write the products in the order in which they are given.)

    In terms of their chemical formulas and physical states, the products can be written as the following.
    I2(g) + H2O(g)

  • Step 5 of 6

    Use the proper notation to write a chemical equation by placing the group of reactants (from Step 2) to the left of an arrow (→) symbol and the group of products (from Step 4) to the right of the arrow symbol. The arrow symbol should point toward the products.

    The chemical equation for the reaction (without coefficients) is the following. (Include states-of-matter under the given conditions in your answer.)
    HI(g) + O2(g) → I2(g) + H2O(g)

  • Step 6 of 6

    Use the proper notation to show the chemical equation by inserting coefficients in front of each reactant and product to match the molar amounts indicated in the reaction description. If the molar amount is 1, show the chemical formula with no coefficient.

    The chemical equation for the reaction (with coefficients) is the following. (Include states-of-matter under the given conditions in your answer.)
    4 HI(g) + O2(g) → 2 I2(g) + 2 H2O(g)

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Sodium azide, NaN3, is the primary gas-forming compound present in many car airbag systems. It is synthesized by reacting sodium amide, NaNH2, with the gas nitrous oxide, N2O, according to the following equation.
2 NaNH2 + N2O → NaN3 + NaOH + NH3
A sample of sodium azide is prepared by reacting 20.22 g NaNH2 with 11.48 g N2O.
(a) What is the limiting reagent?

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NaNH2

(b) How much sodium azide is produced?
WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

16.85

 g NaN3




Solution or Explanation

  • Step 1 of 6

    Calculate the molar masses of the given substances. (Enter an unrounded value.)

    The formula mass of sodium amide, NaNH2, is the following.
    NaNH2 = (1)(average atomic mass of Na) + (1)(average atomic mass of N)
    + (2)(average atomic mass of H)
     = (1)(22.99 amu) + (1)(14.01 amu) + (2)(1.008 amu)
     = 39.016 amu
    The formula mass of nitrous oxide, N2O, is the following.
    N2O = (2)(average atomic mass of N) + (1)(average atomic mass of O)
     = (2)(14.01 amu) + (1)(16.00 amu)
     = 44.02 amu
    Formula mass in amu equals molar mass in g/mol, so the molar mass of NaNH2 is 39.016 g/mol and the molar mass of N2O is 44.02 g/mol.

  • Step 2 of 6

    Calculate the molar mass of the desired substance. (Enter an unrounded value.)

    The desired substance is sodium azide, NaN3. The formula mass of NaN3 is the following.
    NaN3 = (1)(average atomic mass of Na) + (3)(average atomic mass of N)
     = (1)(22.99 amu) + (3)(14.01 amu)
     = 65.02 amu
    Formula mass in amu equals molar mass in g/mol, so the molar mass of NaN3 is 65.02 g/mol.

  • Step 3 of 6

    Determine the stoichiometric ratios of the reactants to the desired substance.

    2 NaNH2 + N2O → NaN3 + NaOH + NH3
    The ratio of NaNH2 to NaN3 is 1 mol NaN3:2 mol NaNH2, and the ratio of N2O to NaN3 is 1 mol NaN3:1 mol N2O.

  • Step 4 of 6

    Calculate the amount of product that would be produced by completely reacting each quantity of reactant.

    20.22 g NaNH2×
    1 mol NaNH2
    39.016 g NaNH2
    		×
    1 mol NaN3
    2 mol NaNH2
    		×
    65.02 g NaN3
    1 mol NaN3
    		 = 16.85 g NaN3
    11.48 g N2O×
    1 mol N2O
    44.02 g N2O
    		×
    1 mol NaN3
    1 mol N2O
    		×
    65.02 g NaN3
    1 mol NaN3
    		 = 16.96 g NaN3
    The answer in each case has four significant figures to match the precision of the least precise values.

  • Step 5 of 6

    Identify the limiting reactant.

    Less NaN3 is produced when limiting reagent is used completely; therefore, the limiting reagent is the following.
    NaNH2

  • Step 6 of 6

    Summarize the answer.

    (a) The limiting reagent is the following.
    NaNH2

    (b) 16.85 g NaN3 will form.

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