### StoichiometryMass-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)

Problem #15: What mass of water is needed to react completely with 108.0 g of oxygen diflouride?
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 ---> 18S8 + SO2 + 2H2O
2H2S + SO2 ---> 38S8 + 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 ---> 12S8 + 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 + 72O2 ---> 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
5) Here's a slightly different approach:
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