Limiting Reagent Problems

#11 - 20

Limiting Reagent Problems #1-10 | Limiting reagent tutorial | Stoichiometry Menu |

**Problem #11:** The equation for the reduction of iron ore in a blast furnace is given below. (a) How many kilograms of iron can be produced by the reaction of 7.00 kg of Fe_{2}O_{3} and 3.00 kg of CO? (b) How many kilograms of the excess reagent remains after reaction has ceased?

Fe_{2}O_{3}+ 3CO ---> 2Fe + 3CO_{2}

**Solution to a:**

1) Determine the limiting reagent:

Fe_{2}O_{3}⇒ 7000 g / 159.687 g/mol = 43.836 mol

CO ⇒ 3000 g / 28.01 g/mol = 107.105 molFe

_{2}O_{3}⇒ 43.836 mol / 1 mol = 43.836

CO ⇒ 107.105 mol / 3 mo = 35.702CO is the limiting reagent.

2) Use the CO : Fe molar ratio:

3 is to 2 as 107.105 mol is to xx = 71.403 mol of Fe produced

3) Convert to kilograms of Fe:

71.403 mol times 55.845 g/mol = 3987.52 gto three sig figs this is 3.99 kg of iron

**Solution to b:**

1) Use Fe_{2}O_{3} : CO molar ratio

1 is to 3 as x is to 107.105 molx = 35.702 mol of Fe

_{2}O_{3}consumed

2) Determine mass remaining:

35.703 mol times 159.687 g/mol = 5701 g consumed7000 g − 5701 g = 1299 g

to three sig figs, this is 1.3 kg

**Problem #12:** The reaction of 4.25 g of Cl_{2} with 2.20 g of P_{4} produces 4.28 g of PCl_{5}. What is the percent yield?

**Solution:**

1) First, a balanced chemical equation:

P_{4}+ 10Cl_{2}---> 4PCl_{5}

2) Get moles, then limiting reagent:

P_{4}⇒ 2.20 g / 123.896 g/mol = 0.0177568 mol

Cl_{2}⇒ 4.25 g / 70.906 g/mol = 0.0599385 molP

_{4}⇒ 0.0177568 / 1 = 0.0177568

Cl_{2}⇒ 0.0599385 / 10 = 0.00599385Cl

_{2}is the limiting reagent.

3) How many grams of PCl_{5} are produced?

Use Cl_{2}: PCl_{5}molar ratio of 10 : 4 (or 5 : 2, if you prefer)5 is to 2 as 0.0599385 is to x

x = 0.0239754 mol of PCl

_{5}produced0.0239754 mol times 208.239 g/mol = 4.99 g ( to three sig figs)

4) Determine percent yield:

(4.28 g / 4.99 g) x 100 = 85.8%

Notice how asking about percent yield (oh, so innocuous!) forces you to go through an entire limiting reagent calculation first.

**Problem #13:** 35.5 g SiO_{2} and 66.5 g of HF react to yield 45.8 g H_{2}SiF_{6} in the folowing equation:

SiO_{2}(s) + 6 HF(aq) ---> H_{2}SiF_{6}(aq) + 2 H_{2}O(l)

(a) How much mass of the excess reactant remains after reaction ceases?

(b) What is the theoretical yield of H_{2}SiF_{6} in grams?

(c) What is the percent yield?

**Solution to (a):**

1) Must determine limiting reagent first (even is it not asked for in the question):

SiO_{2}⇒ 35.5 g / 60.084 g/mol = 0.59084 mol

HF ⇒ 66.5 g / 20.0059 g/mol = 3.324 molSiO

_{2}⇒ 0.59084 mol / 1 mol = 0.59

HF ⇒ 3.324 mol / 6 mol = 0.554HF is limiting.

2) Determine how much SiO_{2} remains:

The SiO_{2}: HF molar ratio is 1 : 61 is to 6 as x is to 3.324 mol

x = 0.554 mol of SiO

_{2}used up0.59084 mol minus 0.554 mol = 0.03684 mol of SiO

_{2}remains0.03684 mol times 60.084 g/mol = 2.21 g (to three sig figs)

**Solution to (b):**

There are 0.59084 mol of SiO_{2}SiO

_{2}: H_{2}SiF_{6}molar ratio is 1 : 1therefore, 0.59084 mol of H

_{2}SiF_{6}produced0.59084 mol times 144.0898 g/mol = 85.1 g (to three sig figs)

**Solution to (c):**

(45.8 g / 85.1 g) times 100 = 53.8%

**Problem #14:** Gaseous ethane reacts with gaseous dioxygen to produce gaseous carbon dioxide and gaseous water.

(a) Suppose a chemist mixes 13.8 g of ethane and 45.8 g of dioxygen. Calculate the theoretical yield of water.(b) Suppose the reaction actually produces 14.2 grams of water . Calculate the percent yield of water.

**Solution to (a):**

1) Write the balanced equation:

2C_{2}H_{6}+ 7O_{2}---> 4CO_{2}+ 6H_{2}O

2) Determine limiting reagent:

C_{2}H_{6}⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol

O_{2}⇒ 45.8 g / 31.9988 g/mol = 1.4313 molC

_{2}H_{6}⇒ 0.45894 / 2 = 0.22947

O_{2}⇒ 1.4313 / 7 = 0.20447Oxygen is limiting.

3) Determine theoretical yield of water:

The oxygen : water molar ratio is 7 : 67 is to 6 as 1.4313 mol is to x

x = 1.2268286 mol of water

4) Convert moles of water to grams:

1.2268286 mol times 18.015 g/mol = 22.1 g (to three sig figs)

**Solution to (b):**

14.2 g / 22.1 g = 64.2%

**Problem #15:** A 0.972-g sample of a CaCl_{2} **⋅** 2H_{2}O and K_{2}C_{2}O_{4} **⋅** H_{2}O solid salt mixture is dissolved in 150 mL of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl_{2} **⋅** 2H_{2}O

(a) How many grams of the excess reactant, K_{2}C_{2}O_{4}⋅H_{2}O, reacted in the mixture?(b) What is the percent by mass of CaCl

_{2}⋅2H_{2}O?(c) How many grams of the K

_{2}C_{2}O_{4}⋅H_{2}O in the salt mixture remain unaffected?

**Solution to (a):**

1) Write the balanced chemical reaction:

CaCl_{2}+ K_{2}C_{2}O_{4}---> CaC_{2}O_{4}+ 2KCl

2) Determine moles of calcium oxalate that precipitated:

0.375 g / 128.096 g/mol = 0.0029275 mol

3) Determine moles, then grams of potassium oxalate:

The K_{2}C_{2}O_{4}: CaC_{2}O_{4}mole ratio is 1:10.0029275 mol of potassium oxalate monohydrate reacted

0.0029275 mol times 184.229 g/mol = 0.53933 g

To three sig figs, 0.539 g

**Solution to (b):**

1) Determine moles, then grams of calcium chloride that reacted:

The CaCl_{2}: CaC_{2}O_{4}mole ratio is 1:10.0029275 mol of calcium chloride dihydrate reacted

0.0029275 mol times 147.0136 g/mol = 0.43038 g

To three sig figs, 0.430 g

2) Determine mass percent of calcium chloride:

0.430 g / 0.972 g = 44.24%

**Solution to (c):**

1) Determine total mass that reacted:

0.430 g + 0.539 g = 0.969 g

2) Determine mass of excess reactant that remains:

0.972 g minus 0.969 g = 0.003 g

**Problem #16:** The reaction of 15.0 g C_{4}H_{9}OH, 22.4 g NaBr, and 32.7 g H_{2}SO_{4} yields 17.1 g C_{4}H_{9}Br in the reaction below:

C_{4}H_{9}OH + NaBr + H_{2}SO_{4}---> C_{4}H_{9}Br + NaHSO_{4}+ H_{2}O

Determine:

(a) the theoretical yield of C_{4}H_{9}Br

(b) the actual percent yield of C_{4}H_{9}Br

(c) the masses of leftover reactants, if any

**Solution to (a):**

1) Determine the limiting reagent bewteen the first two reagents (the third reagent will be dealt with in step 2):

C_{4}H_{9}OH ⇒ 15.0 g / 74.122 g/mol = 0.202369 mol

NaBr ⇒ 22.4 g / 102.894 g/mol = 0.217700 molC

_{4}H_{9}OH ⇒ 0.202369 / 1 =

NaBr ⇒ 0.217700 / 1 =Between these two reactants, C

_{4}H_{9}OH is limiting.

2) Compare C_{4}H_{9}OH to H_{2}SO_{4} to determine which is limiting:

C_{4}H_{9}OH ⇒ 0.202369 mol

H_{2}SO_{4}⇒ 32.7 g / 98.0768 g/mol = 0.333412 molC

_{4}H_{9}OH ⇒ 0.202369 / 1 =

H_{2}SO_{4}⇒ 0.333412 / 1 =Between these two reactants, C

_{4}H_{9}OH is limiting.

Overall, the above process shows that the limiting reagent for the entire reaction is C_{4}H_{9}OH.

3) Determine theoretical yield of C_{4}H_{9}Br:

There is a 1:1 molar ratio between C_{4}H_{9}OH and C_{4}H_{9}Br.This means 0.202369 mol of C

_{4}H_{9}Br is produced.0.202369 mol times 137.019 g/mol = 27.7 g (to three sig figs)

**Solution to (b):**

17.1 g / 27.7 g = 61.7%

**Solution to (c):**

1) Due to the 1:1 molar ratio:

0.202369 mol of NaBr is used up.

2) Therefore:

0.217700 minus 0.202369 = 0.015331 mol of NaBr remains.

The solution for sulfuric acid follows the same path as for NaBr. Conversion to grams is left to the reader.

**Problem #17:** Ozone (O_{3}) reacts with nitric oxide (NO) dishcarged from jet planes to form oxygen gas and nitrogen dioxide. 0.740 g of ozone reacts with 0.670 g of nitric oxide. Determine the identity and quantity of the reactant supplied in excess.

**Solution:**

1) Wrte the balanced chemical equation:

NO + O_{3}---> NO_{2}+ O_{2}

2) Calculate moles:

NO ⇒ 0.670 g / 30.006 g/mol = 0.0223289 mol

O_{3}⇒ 0.740 g / 47.997 g/mol = 0.0154176 mol

3) Determine limitng reagent:

NO and O_{3}are in a 1:1 molar ratio. O_{3}is limiting, making NO the compound in excess

4) Determine quantity of excess reagent:

Based on the 1:1 ratio, we know 0.0154176 mol of NO is used up. Therefore:0.0223289 mol minus 0.0154176 mol = 0.0069113 mol of NO remainingQuantity means grams:

0.0069113 mol times 30.006 g/mol = 0.207 g (to three significant figures)

**Problem #18:** If 1.24 g of P_{4} reacts with 0.12 g of H_{2}, to give 1.25 g of PH_{3}, determine percent yield.

**Solution:**

1) First, the balanced equation:

^{1}⁄_{4}P_{4}+^{3}⁄_{2}H_{2}---> PH_{3}Decided to do it with fractions.

2) Determine moles of P_{4} and H_{2}:

1.24 g / 123.896 g/mol = 0.01001 mol

0.12 g / 2.016 g/mol = 0.059524 mol

3) Determine the limiting reagent:

0.01001 / 0.25 = 0.04004

0.059524 / 1.5 = 0.039683H

_{2}, by a nose!

4) Determine moles of PH_{3} that can be made from 0.059524 mol of H_{2}:

The molar ratio is 1.5 to 11.5 is to 1 as 0.059524 mol is to x

x = 0.039683 mol

5) Determine mass of PH_{3} (this would be the 100% yield amount):

0.039683 mol times 33.9977 g/mol = 1.35 g (to three sig figs)

6) Percent yield:

(1.25 / 1.35) * 100 = 92.6%

**Problem #19:** What mass of phosphorus (P_{4}) is produced when 41.5 g of calcium phosphate, 26.5 g of silicon diioxide, and 7.80 g of carbon are reacted according to the eqation:

2Ca_{3}(PO_{4})_{2}+ 6SiO_{2}+ 10C ---> P_{4}+ 6CaSiO_{3}+ 10CO

**Solution:**

1) Calculate the moles of each reactant:

Ca_{3}(PO_{4})_{2}---> 41.5 g / 310.174 g/mol = 0.133796 mol

SiO_{2}---> 26.5 g / 60.084 g/mol = 0.44105 mol

carbon ---> 7.80 g / 12.011 g/mol = 0.649405 mol

2) We now write, for each reactant, the ratio of moles present to the coefficient:

Ca_{3}(PO_{4})_{2}---> 0.133796 / 2 = 0.066898

SiO_{2}---> 0.44105 / 6 = 0.07351

carbon ---> 0.649405 / 10 = 0.0649405 <--- smallest value

3) Carbon is the limiting reagent.

4) Determine moles of P_{4} produced:

The C to P_{4}molar ratio is 10 to 110 is to 1 as 0.649405 is to x

x = 0.0649405 mol of P

_{4}

5) Determine mass of P_{4} produced:

(0.0649405 mol) (123.896 g/mo) = 8.045868 gFollowing the rule for rounding with five, 8.04 g.

**Problem #20:** 320 g of sulfur dioxide, 32 g of oxygen, and 64 g water react. (a) which reactant is the limiting reagent and (b) how much H_{2}SO_{4} is produced?

2SO_{2}+ O_{2}+ 2H_{2}O ---> 2H_{2}SO_{4}

Comment: I'm going to solve this problem by doing three mass-mass calculations using SO_{2}, then O_{2}, then H_{2}O.The answer that yields the smallest amount of H_{2}SO_{4} will answer both (a) and (b).

**Solution using SO _{2}:**

determine moles SO_{2}---> 320 g / 64 g/mol = 5.0 moluse 1:1 molar ratio to determine 5.0 mol of H

_{2}SO_{4}produceddetermine mass of H

_{2}SO_{4}---> (5.0 mol) (98 g/mol) = 490 g

**Solution using O _{2}:**

determine moles O_{2}---> 32 g / 32 g/mol = 1.0 moluse 1:2 molar ratio to determine 2.0 mol of H

_{2}SO_{4}produceddetermine mass of H

_{2}SO_{4}---> (2.0 mol) (98 g/mol) = 196 g

**Solution using H _{2}O:**

determine moles H_{2}O ---> 64 g / 18 g/mol = 3.6 moluse 1:1 molar ratio to determine 3.6 mol of H

_{2}SO_{4}produceddetermine mass of H

_{2}SO_{4}---> (3.6 mol) (98 g/mol) = 353 g

The final answer is that O_{2} is the limiting reagent and that 196 g of H_{2}SO_{4} is produced.

I copied this problem from Yahoo Answers. The answerer has a nice explanation on how to determine the limiting reagent. Take a look at it.

Limiting Reagent Problems #1-10 | Limiting reagent tutorial | Stoichiometry Menu |