Limiting Reagent Problems

#11 - 25

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 mol = 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) (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} to 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) (159.687 g/mol) = 5701 g consumed7000 g − 5701 g = 1299 g

to three sig figs, this is 1.30 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}produced(0.0239754 mol) (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 − 0.554 mol = 0.03684 mol of SiO

_{2}remains(0.03684 mol) (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}produced(0.59084 mol) (144.0898 g/mol) = 85.1 g (to three sig figs)

**Solution to (c):**

(45.8 g / 85.1 g) (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) (18.015 g/mol) = 22.1 g (to three sig figs)

**Solution to (b):**

(14.2 g / 22.1 g) (100) = 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 1:1therefore, 0.0029275 mol of potassium oxalate monohydrate reacted

0.0029275 mol) (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 1:1therefore, 0.0029275 mol of calcium chloride dihydrate reacted

(0.0029275 mol) (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)(100) = 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 − 0.969 g = 0.003 gRemember, we were given only the total mass of the mixture. Also, we know the remains of the mixture contain zero calcium chloride, since it has been determined to be the limiting reagent.

**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 = 0.202369

H_{2}SO_{4}---> 0.333412 / 1 = 0.333412Between 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) (137.019 g/mol) = 27.7 g (to three sig figs)

**Solution to (b):**

(17.1 g / 27.7 g) (100) = 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 − 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 − 0.0154176 mol = 0.0069113 mol of NO remainingQuantity means grams:

(0.0069113 mol) (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) (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 an "answers" website. The answerer has a nice explanation on how to determine the limiting reagent. Here it is:

Think of the coefficients as a "recipe" for the reaction. The "recipe" calls for 2 mol SO_{2}, 1 mol O_{2}, and 2 mol H_{2}O. Convert each of the given masses into moles and find out how many "recipes" each substance can make. The substance that makes the fewest "recipes" is the limiting reagent.

One "recipe" is:

2 mol SO_{2}= 2 x 64.1 g = 128 g

1 mol O_{2}= 32.0 g

2 mol H_{2}O = 2 x 18.0 = 36 g

Determine "recipes:"

320 g SO_{2}x (1 recipe / 128 g) = 2.50 recipes

32 g O_{2}x (1 recipe / 32.0 g ) = 1.00 recipes

64 g H_{2}O x (1 recipe / 36 g) = 1.78 recipes32 g O

_{2}is the limiting reagent because it makes the fewest "recipes." You would use the 32 g O_{2}to find the amount of H_{2}SO_{4}produced.

**Problem #21:** Consder the following reaction:

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

If 15.0 g of C_{4}H_{9}OH react with 22.4 g of NaBr and 32.7 g of H_{2}SO_{4} to yield 17.1 g of C_{4}H_{9}Br, what is the percent yield of this reaction?

**Solution:**

1) The equation is balanced as written.

2) Determine moles of each reactant:

C_{4}H_{9}OH ---> 15.0 g / 74.122 gmol = 0.20237 mol

NaBr ---> 22.4 g / 102.894 gmol = 0.2177 mol

H_{2}SO_{4}---> 32.7 g / 98.0768 g/mol = 0.3334 mol

3) Dividing the mole amount of each substance by the respective coefficient of the balanced equation will identify the limiting reagent since it will have the lowest value answer. All the coefficients have a value of 1, therefore it is seen that C_{4}H_{9}OH is the limiting reagent.

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

Since I have already calculated the moles of C_{4}H_{9}OH, that is the value I start with.

0.20237 mol C _{4}H_{9}OH1 mol C _{4}H_{9}Br137.0191 g C _{4}H_{9}Br–––––––––––––––––– x ––––––––––––– x ––––––––––––––– = 27.7286 g C _{4}H_{9}Br1 mol C _{4}H_{9}OH1 mol C _{4}H_{9}Br

5) Determine percent yield:

(17.1 g / 27.7286 g) * 100 = 61.7%

**Problem #22:** Consder the following reaction:

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

Given that 50.0 g of Fe is present initially, determine the limiting reagent and the excess reagent.

**Solution:**

The are no calculations associated with the solution to this problem.Since the amount of oxygen is not stated, the convention used in chemistry is to assume it is present in excess.

This makes the iron the limiting reagent.

**Problem #23:** Determine the limiting and excess reagents for the following reaction:

Mg(OH)_{2}+ 2HCl ---> MgCl_{2}+ H_{2}O

And given the following amounts: 0.0218 mole of Mg(OH)_{2} and 0.0436 mole of HCl

**Solution:**

1) Assume Mg(OH)_{2} is limiting. How much MgCl_{2} is produced?

0.0218 mol Mg(OH) _{2}1 mol MgCl _{2}––––––––––––––––– x ––––––––––– = 0.0218 mol MgCl _{2}1 mol Mg(OH) _{2}

2) Assume HCl is limiting. How much MgCl_{2} is produced?

0.0436 mol HCl 1 mol MgCl _{2}––––––––––––– x –––––––––– = 0.0218 mol MgCl _{2}2 mol HCl

3) The two reactants run out simultaneously. In this case, some teachers (& textbooks) consider there to be two limiting reagents. Other teachers (& textbooks) tend to say there is not a limiting reagent. Both have a point, so the ChemTeam (when he was in the classroom) would point out this situation, say be aware of it in the future and then tell his classes there would not be a question about this point on the test.

**Problem #24:** Someday.

**Problem #25:** Someday.

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