### Worksheet - Limiting Reagent Problems #1 - 15

Problem #1: For the combustion of sucrose:

C12H22O11 + 12O2 ---> 12CO2 + 11H2O

there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

Solution path #1:

1) Calculate moles of sucrose:

10.0 g / 342.2948 g/mol = 0.0292146 mol

2) Calculate moles of oxygen required to react with moles of sucrose:

From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore:

0.0292146 mol times 12 = 0.3505752 mole of oxygen required

3) Determine limiting reagent:

Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol

Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent.

Solution path #2:

1) Calculate moles:

sucrose ⇒ 0.0292146 mol
oxygen ⇒ 0.3125 mol

2) Divide by coefficients of balanced equation:

sucrose ⇒ 0.0292146 mol / 1 mol = 0.0292146
oxygen ⇒ 0.3125 mol / 12 mol = 0.02604

Oxygen is the lower value. It is the limiting reagent.

The second method above will be the preferred method to determine the limiting reagent in the following problems.

Problem #2: Calculate the number of NaBr formula units formed when 50 NBr3 molecules and 57 NaOH formula units react?

2NBr3 + 3NaOH ---> N2 + 3NaBr + 3HOBr

Solution:

Comment: we can treat numbers of molecules or formula units in the exact same manner as we would use moles. Keep in mind that the meaning of one mole is that 6.022 x 1023 of that entity (be it molecules or formula units) is present.

1) Determine limiting reagent:

NBr3 ⇒ 50 "moles" / 2 = 25
NaOH ⇒ 57 "moles" / 3 = 19

NaOH is the lmiting reagent.

Note that there need be no conversion from grams to moles. Discussions of numbers of molecules uses numbers that are directly proportional to the number of moles and do not need to be converted.

2) Use NaOH : NaBr molar ratio:

3 is to 3 as 19 is to x

x = 19 "moles"

Correctly phrased, the answer is 19 formula units.

Problem #3: Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al + 3Cl2 ---> 2AlCl3

How many grams of aluminum chloride could be produced from 34.0 g of aluminum and 39.0 g of chlorine gas?

Solution:

1) Determine the limiting reagent:

Al ⇒ 34.0 g / 26.98 g/mol = 1.2602 mol
Cl2 ⇒ 39.0 g / 70.906 g/mol = 0.5500 mol

Al ⇒ 1.2602 mol / 2 =
Cl2 ⇒ 0.5500 mol / 3 =

Seems pretty obvious that chlorine gas is the limiting reagent.

2) Use Cl2 : AlCl3 molar ratio:

3 is to 2 as 0.5500 mol is to x

x = 0.3667 mol of AlCl3 produced

3) Convert to grams:

0.3667 mol times 133.341 g/mol = 48.9 g (to three sig fig)

Why don't you determine the mass of aluminum that remains after the reaction ceases by using the proper molar ratio?

By the way, you could have done it this way:

48.9 g minus 39.0 g = 9.9 g of Al reacted

34.0 g minus 9.9 g = 24.1 g

It only works this second way if you have mass data on every substance in the reaction. Look back at the first problem in this file and you'll see you can't do it using this second way because you don't know anything about the mass of carbon dioxide produced. In that problem, you have to use the molar ratio way.

Problem #4: Interpret reactions in terms of representative particles, then write balanced chemical equations and compare with your results. Determine limiting and excess reagent and the amount of unreacted excess reactant.

a) 3 atoms of carbon combine with 4 molecules of hydrogen to produce methane (CH4)
b) 7 molecules of hydrogen and 2 molecules of nitrogen gases react to produce ammonia
c) 4 molecules of hydrogen and 5 molecules of chlorine react.

Solution to a:

1) The balanced equation:

C + 2H2 ---> CH4

2) Write the carbon-hydrogen molar ratio:

1 : 2

Remember that this ratio can also be understood in terms of atoms and molecules. Thusly:

one atoms of carbon reacts with two molecules of hydrogen

3) Determine limiting reagent:

carbon ⇒ 3/1 = 3
hydrogen ⇒ 4/2 = 2

Hydrogen is the limiting reagent.

4) Determine amount of carbon consumed:

1 is to 2 as x is to 4

x = 2

5) Determine remaining amount of carbon, the excess reagent:

3 minus 2 = 1 atom of carbon remaining

Answers to b: N2 + 3H2 ---> 2NH3

The molar ratio of importance is nitrogen to hydrogen. It is 1 : 3.

Nitrogen is the limiting reagent.

One molecule of hydrogen remains.

H2 + Cl2 ---> 2HCl

1 : 1. Chlorine in excess by one molecule.

Problem #5: Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?

The unbalanced equation is:

Al2S3 + H2O ---> Al(OH)3 + H2S

Solution:

1) Balance the equation:

Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S

2) Determine moles, then limiting reagent:

Al2S3 ⇒ 316.0 g / 150.159 g/mol = 2.104436 mol
H2O ⇒ 493.0 g / 18.015 g/mol = 27.366 mol

Al2S3 ⇒ 2.104436 / 1 = 2.104436
H2O ⇒ 27.366 / 6 = 4.561

Al2S3 is the limiting reagent.

3) Determine grams of water that react:

The molar ratio to use is 1:6

1 is to 6 as 2.104436 mol is to x

x = 12.626616 mol of water used

12.626616 mol times 18.105 g/mol = 227.4685 g

4) Determine excess:

493.0 g minus 227.46848724 = 265.5 g (to 4 sig figs)

Notice how the question only asks about the excess reagent, but you have to go through the entire set of steps (determine moles, determine limiting reagent, use molar ratio) to get to the answer. Tricky!

Problem #6: In this reaction:

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O

6.088 g CaCO3 reacted with 2.852 g HCl. What mass of CaCO3 remains unreacted?

Solution:

1) Let's verify that the HCl is limiting:

CaCO3 ⇒ 6.088 g / 100.086 g/mol = 0.0608277 mol
HCl ⇒ 2.852 g / 36.461 g/mol = 0.0782206 mol

By inspection, we see that HCl is the limiting reagent. (Mentally divide both values by their respective coefficient from the equation to see this.)

Wouldn't that have been cute if you just assumed the HCl was limiting and the question writer made it a bit of a trick question by making the calcium carbonate limiting?

2) Determine moles, then grams of calcium carbonate used:

1 is to 2 as x is to 0.0782206 mol

x = 0.0391103 mol

0.0391103 mol times 100.086 g/mol = 3.914 g

3) Determine grams of CaCO3 remaining:

6.088 g minus 3.914 g = 2.174 g

Problem #7: How many grams of PF5 can be formed from 9.46 g of PF3 and 9.42 g of XeF4 in the following reaction?

2PF3 + XeF4 ---> 2PF5 + Xe

Solution:

1) Limiting reagent:

PF3 ⇒ 9.46 g / 87.968 g/mol = 0.10754 mol
XeF4 ⇒ 9.42 g / 207.282 g/mol = 0.045445 mol

PF3 ⇒ 0.10754 / 2 = 0.05377
XeF4 ⇒ 0.045445 / 1 = 0.045445

XeF4 is limiting

2) Use XeF4 : PF5 molar ratio:

1 is to 2 as 0.045445 mol is to x

x = 0.090890 mol of PF5 produced

3) Determine grams of PF5:

0.090890 mol times 125.964 g/mol = 11.45 g

Problem #8: How many grams of IF5 would be produced using 44.01 grams of I2O5 and 101.0 grams of BrF3?

6I2O5 + 20BrF3 ---> 12IF5 + 15O2 + 10Br2

Solution:

1) Determine limiting reagent:

I2O5 ⇒ 44.01 g / 333.795 g/mol = 0.1318474 mol
BrF3 ⇒ 101.0 g / 136.898 g/mol = 0.7377756 mol

I2O5 ⇒ 0.1318474 / 6 = 0.02197457
BrF3 ⇒ 0.7377756 / 20 = 0.03688878

I2O5 is limiting.

2) Use I2O5 : IF5 molar ratio:

The ratio is 6 to 12, so I'll use 1 to 2

1 is to 2 as 0.1318474 mol is to x

x = 0.2636948 mol of IF5 produced

3) Convert moles to grams:

0.2636948 mol times 221.89 g/mol = 58.51 g (to 4 sig figs)

Problem #9: 950.0 grams of copper(II) sulfate are reacted with 460.0 grams of zinc metal. (a) What is the theoretical yield of Cu? (b) If 295.8 grams of copper are actually obtained from this reaction, what is the percent yield?

Solution to a:

1) The balanced chemical equation is:

CuSO4 + Zn ---> ZnSO4 + Cu

2) Determine limiting reagent:

CuSO4 ⇒ 950.0 g / 159.607 g/mol = 5.95212 mol
Zn ⇒ 460.0 g / 65.38 g/mol = 7.03579 mol

CuSO4 is limiting.

3) Determine moles, then grams of Cu:

5.95212 mol of Cu is produced (due to the 1 : 1 molar ratio involved)

5.95212 mol times 63.546 g/mol = 378.2 g

Solution to b:

Percent yield is:

295.8 g / 378.2 g = 78.21 %

Problem #10: What weight of each substance is present after 0.4500 g of P4O10 and 1.5000 g of PCl5 are reacted completely?

P4O10 + 6PCl5 ---> 10POCl3

Solution:

1) Determine limiting reagent:

P4O10 ⇒ 0.4500 g / 283.886 g/mol = 0.00158514 mol
PCl5: 1.5000 / 208.239 g/mol = 0.00720326 mol

P4O10 ⇒ 0.00158514 / 1 = 0.00158514
PCl5: 0.00720326 / 6 = 0.00120054

PCl5 is limiting.

2) Determine mass of P4O10 remaining:

Use 1 : 6 molar ratio.

1 is to 6 as x is to 0.00720326 mol

x = 0.00120054 mol of P4O10 remaining

0.00158514 mol minus 0.00120054 mol = 0.0003846 mol

0.0003846 mol times 283.886 g/mol = 0.1092 g

3) Determine mass of POCl3 produced:

Use 6 : 10 molar ratio (or, if you prefer, use 3 : 5).

3 is to 5 as 0.00720326 mol is to x

x = 0.01200543 mol of POCl3 produced

0.01200543 mol times 153.332 g/mol = 1.8408 g

Since PCl5 is limiting, zero grams of it will remain.

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:

(4.25 g / 4.99 g) x 100 = 85.2%

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 / 5 = 0.28626

Ethane is limiting.

3) Determine theoretical yield of water:

The ethane : water molar ratio is 2 : 6 (or 1 : 3, if you prefer)

1 is to 3 as 1.4313 mol is to x

x = 4.2939 mol of water

4) Convert moles of water to grams:

4.2939 mol times 18.015 g/mol = 77.4 g (to three sig figs)

Solution to b:

14.2 g / 77.4 g = 18.3%

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