Mass-Mass Problems

#11 - 25

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Prob #1-10 | Return to Stoichiometry Menu |

**Problem #11:** NH_{3} 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:

4NH_{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(g)

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

1 mol O _{2}4 mol NO 30.006 g NO 6.40 g O _{2}x––––––––– x ––––––– x –––––––––– = 4.80 g NO 31.998 g O _{2}5 mol O _{2}1 mol NO written in one line, it looks like this:

6.40 g O

_{2}x (1 mol O_{2}/ 31.998 g O_{2}) x (4 mol NO / 5 mol O_{2}) 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 O_{2}to moles:6.40 g / 31.998 g/mol = 0.2000125 mol O_{2}(b) use a ratio and proportion involving O

_{2}and NO:

4 mol NO x ––––––– = ––––––– 5 mol O _{2}0.2000125 mol O _{2}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 + N_{2}+ 3O_{2}---> Mg(NO_{3})_{2}The key point will be the 3:1 ratio between O

_{2}and Mg(NO_{3})_{2}.By the way, the above chemical reaction does not occur in nature, but the coefficients do accurately reflect how much Mg, N

_{2}and O_{2}are needed to make magnesium nitrate.

2) Determine how many moles are in the 20.00 g of O_{2}:

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(NO

_{3})_{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:

2H_{2}O ---> 2H_{2}+ O_{2}

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 xx = 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) + 3CuSO_{4}(aq) ---> 3Cu(s) + Al_{2}(SO_{4})_{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 xx = 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 Al_{2}(SO_{4})_{3} molar ratio:

2 is to 1 as 0.956264 mol is to xx = 0.478132 mol of Al

_{2}(SO_{4})_{3}

2) Determine grams of aluminum sulfate:

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

OF_{2}(g) + H_{2}O(ℓ) ---> 2HF(aq) + O_{2}(g)

**Solution:**

1) Determine moles of OF_{2}:

108.0 g / 53.995 g/mol = 2.000 mol

2) Use OF_{2} to H_{2}O molar ratio to determine moles of H_{2}O:

1 is to 1 as 2.000 mol is to xx = 2.000 mole of H

_{2}O

3) Determine grams of H_{2}O:

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

**Problem #16:** Based on this balanced equation:

10Li + N_{2}F_{4}---> 4LiF + 2Li_{3}N

(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 xx = 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 10^{23} of the entities -- be they atoms, molecules or formula units -- under discussion.)

**Solution to b:**

1) In every one molecule of N_{2}F_{4}, there are two atoms of N. We need to know how many molecules of N_{2}F_{4} are present:

570 atoms of N divided by 2 atoms of N per one molecule of N_{2}F_{4}= 285 molecules of N_{2}F_{4}

2) Use the N_{2}F_{4} to LiF ratio to determine formula units of LiF

1 is to 4 as 285 is to xx = 1140 formula units of LiF

Notice the extra step. The N_{2}F_{4} reacts as a molecular unit, not individual atoms of N. Hence, I needed to determine how many molecules of N_{2}F_{4} 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 Li_{2}, I would have divided 570 by 2. if it had been Li_{4}, 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 + 2H_{2}O ---> 2MOH + H_{2}

2) Determine moles of H_{2} produced:

1.19 g / 2.016 g/mol = 0.59028 mol

3) Moles M required:

Molar ratio between M and H_{2}is 2:12 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/molM is potassium.

**Problem #18:** Mg + 2HCl ---> MgCl_{2} + H_{2}

(a) How many grams of magnesium (Mg) are needed to produce 100.0 grams of hydrogen (H_{2})?

(b) How many grams of hydrogen chloride (HCl) is needed to produce 200.0 g of hydrogen (H_{2})?

(c) If 500. g of magnesium chloride (MgCl_{2}) are produced in the above reaction, how many grams of hydrogen (H_{2}) would be produced?

**Solution to (a):**

1 mol H _{2}1 mol Mg 24.305 g Mg 100.0 g H _{2}x––––––––– x ––––––– x –––––––––– = 1206 g Mg (to four sig figs) 2.016 g H _{2}1 mol H _{2}1 mol Mg

**Solution to (b):**

1 mol H _{2}2 mol HCl 36.4609 g HCl 200.0 g H _{2}x––––––––– x ––––––––––– x –––––––––––– = 7234 g HCl 2.016 g H _{2}1 mol MgCl _{2}2 mol HCl

**Solution to (c):**

1 mol MgCl _{2}1 mol H _{2}2.016 g H _{2}500.0 g MgCl _{2}x–––––––––––– x –––––––––– x ––––––––– = 10.59 g H _{2}95.211 g MgCl _{2}1 mol MgCl _{2}1 mol H _{2}

**Problem #19:** P_{4}(s) + 5O_{2}(g) ---> P_{4}O_{10}(g)

(a) How many grams of phosphorus(V) oxide (P_{4}O_{10}) are produced if you burn 50.0 grams of phosphorus with sufficient oxygen (O_{2})?

(b) How many grams of oxygen would be needed in part (a)?

(c) If 400. grams of phosphorus(V) oxide (P_{4}O_{10}) 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 P_{4}O_{10} produced:

1 1000 g ––––––– = ––––––– 1 x x = 0.403564 mol (of P

_{4}O_{10}, NOT P_{4})

3) Determine grams of P_{4}O_{10} produced:

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

**Solution to (b):**

1 mol P _{4}5 mol O _{2}32.00 g O _{2}50.0 g P _{4}x–––––––––– x –––––––– x ––––––––– = 64.57 g H _{2}123.896 g P _{4}1 mol P _{4}1 mol O _{2}

**Solution to (c):**

1) Determine moles of P_{4}O_{10} produced:

400. g / 283.886 g/mol = 1.4090163 mol

2) The molar ratio between P_{4} and P_{4}O_{10} is 1:1

Therefore, 1.4090163 mol of P_{4}is required.

3) Determine grams of P_{4} required:

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

2H_{2}S + 2O_{2}--->^{1}⁄_{8}S_{8}+ SO_{2}+ 2H_{2}O

2H_{2}S + SO_{2}--->^{3}⁄_{8}S_{8}+ 2H_{2}O

What mass of sulfur can be produced from 48.0 grams of O_{2}?

**Solution:**

1) Let's add the two equations together:

4H_{2}S + 2O_{2}--->^{1}⁄_{2}S_{8}+ 4H_{2}OThere is a 4:1 molar ratio between O

_{2}and S_{8}

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 xx = 0.375 mol of S

_{8}

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.

2KClO_{3}---> KCl + 3O_{2}

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 O _{2}32.00 g 240. g x ––––––– x ––––––– x ––––––– = 94.0 g O _{2}122.5 g 2 mol KClO _{3}1 mol

2) We examine the answer choices, looking for:

240, 3, and 32 in the numerator

122.5 and 2 in the denominatorWe find that answer choice (b) fits our needs.

**Problem #22:** A mixture contains no fluorine compounds except for methyl fluoroacetate, FCH_{2}COOCH_{3}. When chemically treated, all the fluorine is converted to CaF_{2}. The mass of CaF_{2} was found to be 12.10 g. Determine the mass of methyl fluoroacetate in the original mixture.

**Solution:**

1) Determine the moles of CaF_{2}:

12.10 g / 78.074 g/mol = 0.15498 mol

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

CaH_{2}+ 2FCH_{2}COOCH_{3}---> CaF_{2}+ 2CH_{3}COOCH_{3}Note that two methyl fluoroacetate are consumed for every one CaF

_{2}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 CaF_{2} 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) + H_{2}SO_{4}(aq) ---> CuSO_{4}(aq) + H_{2}O(ℓ)

Zn(s) + CuSO_{4}(aq) ---> ZnSO_{4}(aq) + Cu(s)

**Solution:**

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

Zn(s) + CuO(s) + H_{2}SO_{4}(aq) ---> H_{2}O(ℓ) + ZnSO_{4}(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 CO_{2} gas is produced by the combustion of 162.9 g of a mixture that is 61.1% CH_{4} and 38.9% C_{3}H_{8} by mass?

**Solution:**

1) Determine mass of each hydrocarbon:

mass CH_{4}---> (162.9 g) (0.611) = 99.5319 g

mass C_{3}H_{8}---> 162.9 g − 99.5319 g = 63.3681 g

2) Write the combustion reaction for methane:

CH_{4}+ 2O_{2}---> CO_{2}+ 2H_{2}OThere 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 CO_{2} was produced.

5) Determine the mass of CO_{2} produced:

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

6) Write the combustion equation for propane:

C_{3}H_{8}+ 5O_{2}---> 3CO_{2}+ 4H_{2}OThere 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 CO_{2} are produced:

(1.437042 mol) (3) = 4.311126 mol

9) Determine grams of CO_{2}:

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

10) Total CO_{2} produced:

273 g + 190. g = 463 g

**Problem #25:** Ethane gas (C_{2}H_{6}) 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:

C_{2}H_{6}+^{7}⁄_{2}O_{2}---> 2CO_{2}+ 3H_{2}OThere 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 CO_{2} produced:

2 is to 3 as x is to 0.4440744 molx = 0.2960496 mol

4) Determine mass of CO_{2}:

(0.2960496 mol) (44.009 g/mol) = 13.0 g5) Here's a slightly different approach:

Convert 2CO_{2}(two moles of CO_{2}) and 3H_{2}O (three moles of H_{2}O) to grams:2CO_{2}---> (44.009 g/mol) (2 mol) = 88.018 g

3H_{2}O ---> (18.015 g/mol) (3 mol) = 54.045 gUse a ratio and proportion, but in grams, not moles:

88.018 x ––––––– = ––––––– 54.045 8.00 x = 13.0 g

Ten Examples | |

Prob #1-10 | Return to Stoichiometry Menu |