Worksheet - Limiting Reagent Problems #11 - 20

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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 Fe2O3 and 3.00 kg of CO? (b) How many kilograms of the excess reagent remains after reaction has ceased?

Fe2O3 + 3 CO ---> 2 Fe + 3 CO2

Solution to a:

1) Determine the limiting reagent:

Fe2O3 ⇒ 7000 g / 159.687 g/mol = 43.836 mol
CO ⇒ 3000 g / 28.01 g/mol = 107.105 mol

Fe2O3 ⇒ 43.836 mol / 1 mol = 43.836
CO ⇒ 107.105 mol / 3 mo = 35.702

CO is the limiting reagent.

2) Use the CO : Fe molar ratio:

3 is to 2 as 107.105 mol is to x

x = 71.403 mol of Fe produced

3) Convert to kilograms of Fe:

71.403 mol times 55.845 g/mol = 3987.52 g

to three sig figs this is 3.99 kg of iron

Solution to b:

1) Use Fe2O3 : CO molar ratio

1 is to 3 as x is to 107.105 mol

x = 35.702 mol of Fe2O3 consumed

2) Determine mass remaining:

35.703 mol times 159.687 g/mol = 5701 g consumed

7000 g minus 5701 g = 1299 g

to three sig figs, this is 1.3 kg


Problem #12: The reaction of 4.25 g of Cl2 with 2.20 g of P4 produces 4.28 g of PCl5. What is the percent yield?

Solution:

1) First, a balanced chemical equation:

P4 + 10Cl2 ---> 4PCl5

2) Get moles, then limiting reagent:

P4 ⇒ 2.20 g / 123.896 g/mol = 0.0177568 mol
Cl2 ⇒ 4.25 g / 70.906 g/mol = 0.0599385 mol

P4 ⇒ 0.0177568 / 1 = 0.0177568
Cl2 ⇒ 0.0599385 / 10 = 0.00599385

Cl2 is the limiting reagent.

3) How many grams of PCl5 are produced?

Use Cl2 : PCl5 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 PCl5 produced

0.0239754 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 SiO2 and 66.5 g of HF react to yield 45.8 g H2SiF6 in the folowing equation:

SiO2(s) + 6 HF(aq) ---> H2SiF6(aq) + 2 H2O(l)

a. How much mass of the excess reactant remains after reaction ceases?
b. What is the theoretical yield of H2SiF6 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):

SiO2 ⇒ 35.5 g / 60.084 g/mol = 0.59084 mol
HF ⇒ 66.5 g / 20.0059 g/mol = 3.324 mol

SiO2 ⇒ 0.59084 mol / 1 mol = 0.59
HF ⇒ 3.324 mol / 6 mol = 0.554

HF is limiting.

2) Determine how much SiO2 remains:

The SiO2 : HF molar ratio is 1 : 6

1 is to 6 as x is to 3.324 mol

x = 0.554 mol of SiO2 used up

0.59084 mol minus 0.554 mol = 0.03684 mol of SiO2 remains

0.03684 mol times 60.084 g/mol = 2.21 g (to three sig figs)

Solution to b:

There are 0.59084 mol of SiO2

SiO2 : H2SiF6 molar ratio is 1 : 1

therefore, 0.59084 mol of H2SiF6 produced

0.59084 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:

2C2H6 + 7O2 ---> 4CO2 + 6H2O

2) Determine limiting reagent:

C2H6 ⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol
O2 ⇒ 45.8 g / 31.9988 g/mol = 1.4313 mol

C2H6 ⇒ 0.45894 / 2 = 0.22947
O2 ⇒ 1.4313 / 7 = 0.20447

Oxygen is limiting.

3) Determine theoretical yield of water:

The oxygen : water molar ratio is 7 : 6

7 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 CaCl2 2H2O and K2C2O4 H2O 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 CaCl2 2H2O

a. How many grams of the excess reactant, K2C2O4 H2O, reacted in the mixture?

b. What is the percent by mass of CaCl2 2H2O?

c. How many grams of the K2C2O4 H2O in the salt mixture remain unaffected?

Solution to a:

1) Write the balanced chemical reaction:

CaCl2 + K2C2O4 ---> CaC2O4 + 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 K2C2O4 : CaC2O4 mole ratio is 1:1

0.0029275 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 CaCl2 : CaC2O4 mole ratio is 1:1

0.0029275 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 C4H9OH, 22.4 g NaBr, and 32.7 g H2SO4 yields 17.1 g C4H9Br in the reaction below:

C4H9OH + NaBr + H2SO4 ---> C4H9Br + NaHSO4 + H2O

Determine:

(a) the theoretical yield of C4H9Br
(b) the actual percent yield of C4H9Br
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):

C4H9OH ⇒ 15.0 g / 74.122 g/mol = 0.202369 mol
NaBr ⇒ 22.4 g / 102.894 g/mol = 0.217700 mol

C4H9OH ⇒ 0.202369 /1 =
NaBr ⇒ 0.217700 / 1 =

Between these two reactants, C4H9OH is limiting.

2) Compare C4H9OH to H2SO4 to determine which is limiting:

C4H9OH ⇒ 0.202369 mol
H2SO4 ⇒ 32.7 g / 98.0768 g/mol = 0.333412 mol

C4H9OH ⇒ 0.202369 /1 =
H2SO4 ⇒ 0.333412 / 1 =

Between these two reactants, C4H9OH is limiting.

Overall, the above process shows that the limiting reagent for the entire reaction is C4H9OH.

3) Determine theoretical yield of C4H9Br:

There is a 1:1 molar ratio between C4H9OH and C4H9Br.

This means 0.202369 mol of C4H9Br 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 (O3) 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 + O3 ---> NO2 + O2

2) Calculate moles:

NO ⇒ 0.670 g / 30.006 g/mol = 0.0223289 mol
O3 ⇒ 0.740 g / 47.997 g/mol = 0.0154176 mol

3) Determine limitng reagent:

NO and O3 are in a 1:1 molar ratio. O3 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 remaining

Quantity means grams:

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

Problem #18: If 1.24 g of P4 reacts with 0.12 g of H2, to give 1.25 g of PH3, determine percent yield.

Solution:

1) First, the balanced equation:

(1/4)P4 + (3/2)H2 ---> PH3

Decided to do it with fractions.

2) Determine moles of P4 and H2:

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 = 0.039683

H2, by a nose!

4) Determine moles of PH3 that can be made from 0.059524 mol of H2:

The molar ratio is 1.5 to 1

1.5 is to 1 as 0.059524 mol is to x

x = 0.039683 mol

5) Determine mass of PH3 (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%

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