Stoichiometry
Mass-Mass Problems
#11 - 25

Problem #11: NH₃ chemically reacts with oxygen gas to produce nitric oxide and water. What mass of nitric oxide is produced by the reaction of 6.40 g of oxygen gas?

Solution:

1) Write the balanced chemical equation:

4NH₃(g) + 5O₂(g) ---> 4NO(g) + 6H₂O(g)

2) The solution, done in the style of dimensional analysis:

	1 mol O2		4 mol NO		30.006 g NO
6.40 g O2 x	–––––––––	x	–––––––	x	––––––––––	= 4.80 g NO
	31.998 g O2		5 mol O2		1 mol NO

written in one line, it looks like this:

6.40 g O₂ x (1 mol O₂ / 31.998 g O₂) x (4 mol NO / 5 mol O₂) x (30.006 g NO / 1 mol NO) = 4.80 g NO

3) Written in individual steps, the solution looks like this:

(a) convert grams of O₂ to moles:

6.40 g / 31.998 g/mol = 0.2000125 mol O₂

(b) use a ratio and proportion involving O₂ and NO:

4 mol NO		x
–––––––	=	–––––––
5 mol O2		0.2000125 mol O2

x = 0.16001 mol NO

(c) convert moles of NO to grams:

(0.16001 mol) (30.006 g/mol) = 4.80 g NO

Problem #12: How many grams of magnesium nitrate can be formed from 20.00 g of oxygen gas?

Solution:

1) Let us write a balanced chemical equation:

Mg + N₂ + 3O₂ ---> Mg(NO₃)₂

The key point will be the 3:1 ratio between O₂ and Mg(NO₃)₂.

By the way, the above chemical reaction does not occur in nature, but the coefficients do accurately reflect how much Mg, N₂ and O₂ are needed to make magnesium nitrate.

2) Determine how many moles are in the 20.00 g of O₂:

20.00 g / 31.9988 g/mol = 0.62502344 mol

3) We will now use the 3:1 ratio:

3		0.62502344 mol
–––	=	–––––––––––––
1		x

x = 0.208341 mol of Mg(NO₃)₂

4) Determine the mass of magnesium nitrate required:

(148.313 g/mol) (0.208341 mol) = 30.90 g (to four sig figs)

Problem #13: Water decomposes. How many moles of oxygen can be produced from 2.40 g of water at STP?

Solution:

1) Write a balanced chemical equation:

2H₂O ---> 2H₂ + O₂

2) Determine moles of water:

2.40 g / 18.015 g/mol = 0.13322 mol

3) The water to oxygen molar ratio is 2:1. Determine moles of oxygen produced:

2 is to 1 as 0.13322 mol is to x

x = 0.0666 mol (to three sig figs)

Note that, since this is a mass-based problem, there is no need to use STP anywhere in the calculation.

Problem #14: Given this equation:

2Al(s) + 3CuSO₄(aq) ---> 3Cu(s) + Al₂(SO₄)₃(aq).

(a) From 25.80 g of aluminum, how many grams of copper are produced?
(b) How many moles of aluminum sulfate are produced?

Solution to (a):

1) Determine moles of Al:

25.80 g / 26.98 g/mol = 0.956264 mol (I kept a few gaurd digits)

2) Use Al:Cu molar ratio to determine moles of Cu produced:

2 is to 3 as 0.956264 mol is to x

x = 1.434396 mol of Cu

3. Determine grams of copper:

1.434396 mol times 63.546 g/mol = 91.15 g (to four sig figs)

Solution to (b):

1) Use Al to Al₂(SO₄)₃ molar ratio:

2 is to 1 as 0.956264 mol is to x

x = 0.478132 mol of Al₂(SO₄)₃

2) Determine grams of aluminum sulfate:

0.478132 mol times 342.147 g/mol = 163.6 g (to four sig figs)

OF₂(g) + H₂O(ℓ) ---> 2HF(aq) + O₂(g)

Solution:

1) Determine moles of OF₂:

108.0 g / 53.995 g/mol = 2.000 mol

2) Use OF₂ to H₂O molar ratio to determine moles of H₂O:

1 is to 1 as 2.000 mol is to x

x = 2.000 mole of H₂O

3) Determine grams of H₂O:

2.000 mol times 18.015 g/mol = 36.03 g (to four sig figs)

Problem #16: Based on this balanced equation:

10Li + N₂F₄ ---> 4LiF + 2Li₃N

(a) Calculate the formula units of LiF formed when 570 atoms of Li are reacted.
(b) Calculate the formula units of LiF formed, if 570 atoms of N are used in the reaction along with sufficient Li.

Solution to a:

Use the Li:LiF molar ratio to determine formula units of LiF produced:

10 is to 4 as 570 atoms is to x

x = 228 formula units of LiF

Note: if you wished to reduce the 10:4 ratio to 5:2 before calculating, the 5:2 ratio would lead to the correct answer.

Please be aware that discussing NUMBERS of atoms or formula units is analogous to using MOLES. Remember that using moles is simply a short-hand for discussing how many atoms or formula units are present. (Reminder: one mole contains 6.022 x 10²³ of the entities -- be they atoms, molecules or formula units -- under discussion.)

Solution to b:

1) In every one molecule of N₂F₄, there are two atoms of N. We need to know how many molecules of N₂F₄ are present:

570 atoms of N divided by 2 atoms of N per one molecule of N₂F₄ = 285 molecules of N₂F₄

2) Use the N₂F₄ to LiF ratio to determine formula units of LiF

1 is to 4 as 285 is to x

x = 1140 formula units of LiF

Notice the extra step. The N₂F₄ reacts as a molecular unit, not individual atoms of N. Hence, I needed to determine how many molecules of N₂F₄ were composed of 570 atoms of N.

I did not have to do this in (a) because the Li reacted on the basis of individual atoms of Li. If the formula had been Li₂, I would have divided 570 by 2. if it had been Li₄, I would have divided by 4.

Problem #17: 46.0 g of an alkai metal was reacted with water to form the aqueous metal hydroxide along with 1.19 g of hydrogen gas. Which alkai metal was used?

Solution:

1) Let M be the alkali metal. The chemical reaction is this:

2M + 2H₂O ---> 2MOH + H₂

2) Determine moles of H₂ produced:

1.19 g / 2.016 g/mol = 0.59028 mol

3) Moles M required:

Molar ratio between M and H₂ is 2:1

2 is to 1 as x is to 0.59028

x = 1.18056 mol

4) Determine atomic weight of M:

46.0 g / 1.18056 mol = 39.0 g/mol

M is potassium.

Problem #18: Mg + 2HCl ---> MgCl₂ + H₂

(a) How many grams of magnesium (Mg) are needed to produce 100.0 grams of hydrogen (H₂)?
(b) How many grams of hydrogen chloride (HCl) is needed to produce 200.0 g of hydrogen (H₂)?
(c) If 500. g of magnesium chloride (MgCl₂) are produced in the above reaction, how many grams of hydrogen (H₂) would be produced?

Solution to (a):

	1 mol H2		1 mol Mg		24.305 g Mg
100.0 g H2 x	–––––––––	x	–––––––	x	––––––––––	=  1206 g Mg (to four sig figs)
	2.016 g H2		1 mol H2		1 mol Mg

Solution to (b):

	1 mol H2		2 mol HCl		36.4609 g HCl
200.0 g H2 x	–––––––––	x	–––––––––––	x	––––––––––––	=  7234 g HCl
	2.016 g H2		1 mol MgCl2		2 mol HCl

Solution to (c):

	1 mol MgCl2		1 mol H2		2.016 g H2
500.0 g MgCl2 x	––––––––––––	x	––––––––––	x	–––––––––	=  10.59 g H2
	95.211 g MgCl2		1 mol MgCl2		1 mol H2

Problem #19: P₄(s) + 5O₂(g) ---> P₄O₁₀(g)

(a) How many grams of phosphorus(V) oxide (P₄O₁₀) are produced if you burn 50.0 grams of phosphorus with sufficient oxygen (O₂)?
(b) How many grams of oxygen would be needed in part (a)?
(c) If 400. grams of phosphorus(V) oxide (P₄O₁₀) is needed for another experiment, how much phosphorus would have to be burned?

Solution to (a):

1) Determine moles of phosphorus that burned:

50.0 g / 123.896 g/mol = 0.403564 mol

2) Determine moles of P₄O₁₀ produced:

1		1000 g
–––––––	=	–––––––
1		x

x = 0.403564 mol (of P₄O₁₀, NOT P₄)

3) Determine grams of P₄O₁₀ produced:

(0.403564 mol) (283.886 g/mol) = 114 g (to three sig figs)

Solution to (b):

	1 mol P4		5 mol O2		32.00 g O2
50.0 g P4 x	––––––––––	x	––––––––	x	–––––––––	=  64.57 g H2
	123.896 g P4		1 mol P4		1 mol O2

Solution to (c):

1) Determine moles of P₄O₁₀ produced:

400. g / 283.886 g/mol = 1.4090163 mol

2) The molar ratio between P₄ and P₄O₁₀ is 1:1

Therefore, 1.4090163 mol of P₄ is required.

3) Determine grams of P₄ required:

(1.4090163 mol) (123.896 g/mol) = 174.57 g

To three sig figs, this is 174 g (remember the rule of five for rounding)

Problem #20: The Claus reactions, shown below, are used to generate elemental sulfur from hydrogen sulfide.

2H₂S + 2O₂ ---> ¹⁄₈S₈ + SO₂ + 2H₂O
2H₂S + SO₂ ---> ³⁄₈S₈ + 2H₂O

What mass of sulfur can be produced from 48.0 grams of O₂?

Solution:

1) Let's add the two equations together:

4H₂S + 2O₂ ---> ¹⁄₂S₈ + 4H₂O

There is a 4:1 molar ratio between O₂ and S₈

2) Determine moles of oxygen:

48.0 g / 32.0 g/mol = 1.50 mol

3) Using the molar ratio, determine moles of sulfur produced:

4 is to 1 as 1.50 mol is to x

x = 0.375 mol of S₈

4) Determine mass of sulfur produced:

(0.375 mol) (256.52 g/mol) = 96.2 g

Problem #21: In an experiment, potassium chlorate decomposed according to the following chemical equation.

2KClO₃ ---> KCl + 3O₂

If the mass of potassium chlorate was 240. g, which of the following calculations can be used to determine the mass of oxygen gas formed?

(a) (240 x 2 x 32.00) ÷ (122.5 x 3) g
(b) (240 x 3 x 32.00) ÷ (122.5 x 2) g
(c) (240 x 2 x 122.5) ÷ (32.00 x 3) g
(d) (240 x 3 x 122.5) ÷ (32.00 x 2) g

Solution:

1) Set up using dimensional analysis:

	1 mol		3 mol O2		32.00 g
240. g x	–––––––	x	–––––––	x	–––––––	= 94.0 g O2
	122.5 g		2 mol KClO3		1 mol

2) We examine the answer choices, looking for:

240, 3, and 32 in the numerator
122.5 and 2 in the denominator

We find that answer choice (b) fits our needs.

Problem #22: A mixture contains no fluorine compounds except for methyl fluoroacetate, FCH₂COOCH₃. When chemically treated, all the fluorine is converted to CaF₂. The mass of CaF₂ was found to be 12.10 g. Determine the mass of methyl fluoroacetate in the original mixture.

Solution:

1) Determine the moles of CaF₂:

12.10 g / 78.074 g/mol = 0.15498 mol

2) Methyl fluoroacetate is presumed to react with calcium hydride as follows:

CaH₂ + 2FCH₂COOCH₃ ---> CaF₂ + 2CH₃COOCH₃

Note that two methyl fluoroacetate are consumed for every one CaF₂ produced.

By the way, I don't know if the above reaction happens or not. We will assume it does for purposes of this problem.

3) Based on the above molar ratio, we determine that 0.15498 mole of CaF₂ requires 0.30996 mole of methyl fluoroacetate to be consumed.

4) Determine the mass of methyl fluoroacetate that was present in the original mixture:

(0.30996 mol) (92.0685 g/mol) = 28.54 g

Problem #23: Given 0.050 g of copper(II) oxide, how much copper will be produced?

CuO(s) + H₂SO₄(aq) ---> CuSO₄(aq) + H₂O(ℓ)
Zn(s) + CuSO₄(aq) ---> ZnSO₄(aq) + Cu(s)

Solution:

1) Let's add the two reactions together, resulting in:

Zn(s) + CuO(s) + H₂SO₄(aq) ---> H₂O(ℓ) + ZnSO₄(aq) + Cu(s)

The key point is the 1:1 molar ratio between CuO and Cu.

2) Determine moles of CuO:

0.050 g / 79.545 g/mol = 0.00062858 mol

3) Using the 1:1 molar ratio, we determine that 0.00062858 mol of Cu was produced.

4) Determine grams of Cu:

(0.00062858 mol) (63.546 g/mol) = 0.040 g

Problem #24: Hydrocarbon mixtures are used as fuels. What mass of CO₂ gas is produced by the combustion of 162.9 g of a mixture that is 61.1% CH₄ and 38.9% C₃H₈ by mass?

Solution:

1) Determine mass of each hydrocarbon:

mass CH₄ ---> (162.9 g) (0.611) = 99.5319 g
mass C₃H₈ ---> 162.9 g − 99.5319 g = 63.3681 g

2) Write the combustion reaction for methane:

CH₄ + 2O₂ ---> CO₂ + 2H₂O

There is a 1:1 molar ratio between methane and carbon dioxide.

3) Determine moles of methane:

99.5319 g / 16.0426 g/mol = 6.204225 mol

4) The 1:1 molar ratio tells us that 6.204225 mol of CO₂ was produced.

5) Determine the mass of CO₂ produced:

(6.204225 mol) (44.009 g/mol) = 273 g

6) Write the combustion equation for propane:

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

There is a 1:3 molar ratio between propane and carbon dioxide.

7) Determine moles of propane:

63.3681 g / 44.0962 g/mol = 1.437042 mol

8) For every one mole of propane consumed, three moles of CO₂ are produced:

(1.437042 mol) (3) = 4.311126 mol

9) Determine grams of CO₂:

(4.311126 mol) (44.009 g/mol) = 190. g

10) Total CO₂ produced:

273 g + 190. g = 463 g

Problem #25: Ethane gas (C₂H₆) burns in air to form carbon dioxide and water. How many grams of carbon dioxide are produced for each 8.00 grams of water produced?

Solution:

1) Write the combustion reaction:

C₂H₆ + ⁷⁄₂O₂ ---> 2CO₂ + 3H₂O

There is a molar ratio of 2 to 3 between carbon dioxide and water.

2) Determine moles in 8.00 g of water:

8.00 g / 18.015 g/mol = 0.4440744 mol

3) Use the 2:3 molar ratio to determine moles of CO₂ produced:

2 is to 3 as x is to 0.4440744 mol

x = 0.2960496 mol

4) Determine mass of CO₂:

(0.2960496 mol) (44.009 g/mol) = 13.0 g

Convert 2CO₂ (two moles of CO₂) and 3H₂O (three moles of H₂O) to grams:

2CO₂ ---> (44.009 g/mol) (2 mol) = 88.018 g
3H₂O ---> (18.015 g/mol) (3 mol) = 54.045 g

Use a ratio and proportion, but in grams, not moles:

88.018		x
–––––––	=	–––––––
54.045		8.00

x = 13.0 g