Kevin Liu and Radhika Thotakura
Due: January 17, 2013
Chemistry H
Period 3
Limiting Reactant and Percent Yield

Limiting Reactant: The reactant that limits the amount of a product that can
form in a chemical reaction.

Excess Reactant: The substance that is not used up completely in a reaction.
There is excess left of the substance.

Theoretical Yield: (amount calculated) The amount of product that would result
from all of the limiting reactant.

Actual Yield: (given) The amount of the product actually obtained from a
reaction.
Determining Limiting Reactant:
1. Derive and then balance chemical equation.
2. Choose one of the reactants and convert into grams of the other.
o Use factor-label method
3. Compare the converted reactant with the other reactant
o Determine which one is the limiting reactant.
o If the converted reactant is greater than the other, the other reactant is the
limiting reactant.
o If the converted reactant is less than the other, the converted reactant is the
limiting reactant.
Finding Percent Yield
1. Percent Yield = (Actual Yield/Theoretical Yield) x 100
o Actual Yield is the given amount.
o Theoretical Yield is how much product is synthesized in the reaction,
which is the amount you come up with.
THINGS TO REMEMBER:
1. Balance the chemical equation before doing any mathematical calculation.
2. When calculating limiting reactant, don’t forget about diatomic elements (H2, O2,
F2, Br2, I2, N2, Cl2)
3. The amount calculated of the other substance (when calculating limiting reactant
is the amount needed.
4. SIGNIFICANT FIGURES!
5. Label (grams, moles, liters, etc.)
6. Two decimals when giving percentage.
Practice Problems ( SHOW ALL WORK! )
1.
C6H6 + Br2 → C6H5Br + HBr
a. What is the theoretical yield of C6H5Br if 45.2 g of C6H6 reacts with 76.0
g of Br2?
b. If the actual yield of C6H5Br is 64.8 g, what is the percent yield?
2.
Balance the Equation if Unbalanced:
a. H2(g) + N2(g) → NH3(g)
b. If 9.2 g of ammonia was produced from 1.96 g of hydrogen, what was the
percent yield?
3.
What mass of precipitate forms when a solution containing 8.63 g of potassium
sulfate is reacted with a solution containing 17.7 g barium nitrate? (Hint: solubility rules!)
4.
For the unbalanced equation shown below, if the reaction of 132.0 grams of
CaCN2 produces 64.7 grams of NH3, what is the percent yield?
a. CaCN2+ H2O → CaCO3+ NH3
5.
Aluminum dissolves in an aqueous solution of NaOH according to the following
reaction:
a. Balance Equation: NaOH + Al + H2O -----> NaAlO2 + H2
If 95.3 g of NaOH and 59.1 g of Al react:
b. Which is the limiting reactant?
c. How much of the other reactant remains (excess)?
d. What mass of hydrogen is produced?
6.
Fe3O4 + H2 → Fe + H2O
a. Balance the equation.
b. If .123 grams of H2 produces .774 grams of H2O, what is the percent
yield?
Answer Key:
1.
C6H6 + Br2 → C6H5Br + HBr
a. C = 12.01, H = 1.01, Br = 79.90
(45.2 grams C6H6) x (1 mole C6H6/78.12 grams C6H6) x (1 mole Br2/1
mole C6H6) x (159.8 grams Br2/1 mole Br2) = 92. 5 grams Br2
Br2 is the limiting reactant.
(76.0 grams Br2) x (1 mole Br2/159.8 grams Br2) x (1 mole C6H5Br/1
mole Br2) x (157.01 grams C6H5Br/1 mole C6H5Br) = 74.7 grams C6H5Br
b. Percent yield = (64.8 grams/74.7 grams) x 100 = 86.75%
2.
3H2 (g) + N2 (g) → 2NH3 (g)
a. 3H2 (g) + N2 (g) → 2NH3 (g
b. N = 14.01, H = 1.01
(1.96 grams H2) x (1 mole H2/2.02 grams H2) x (2 mole NH3/3 mole H2)
x (17.04 grams NH3/1 mole NH3) = 11.0 grams NH3
Percent Yield = (9.2 grams/11.0 grams) x 100 = 83.64%
3.
Ba(NO3)2(aq) + K2SO4(aq) = KNO3(aq) + BaSO4(s)
BaSO4 is the precipitate
K = 39.10, Ba = 137.33, N = 14.01, O = 16.00, S = 32.07
(8.63 grams K2SO4) x (1 mole K2SO4/174.27 grams K2SO4) x (1 mole
Ba(NO3)2/1 mole K2SO4) x (261.35 grams Ba(NO3)2/1 mole Ba(NO3)2) = 12.9 grams
Ba(NO3)2
K2SO4is the limiting reactant.
(8.63 grams K2SO4) x (1 mole K2SO4/174.27 grams K2SO4) x (1 mole BaSO4/1
mole K2SO4) x (233.4 grams BaSO4/1 mole BaSO4) = 11.6 grams BaSO4
4.
CaCN2+ H2O → CaCO3+ NH3
a. CaCN2+ 3H2O → CaCO3+ 2NH3
b. Ca = 40.08, C = 12.01, N = 14.01, H = 1.01
(132 grams CaCN2) x (1 mole CaCN2/80.11 grams CaCN2) x (2 mole NH3/1
mole CaCN2) x (17.04 grams NH3/1 mole NH3) = 56.2 grams NH3
Percent Yield = (56.2 grams/64.7 grams) x 100 = 86.86%
5.
NaOH + Al + H2O → NaAlO2 + H2
a. 2NaOH + 2Al + 2H2O → 2NaAlO2 + 3H2
b. Na = 22.99, Al = 26.98, H = 1.0, O = 16.00
(95.3 grams NaOH) x (1 mole NaOH/40.0 grams NaOH) x (2 mole Al/2 mole
NaOH) x (26.98 grams Al/1 mole Al) = 64.3 grams Al
Al is the limiting reactant.
c. (59.1 grams Al) x (1 mole Al/26.98 grams Al) x (2 mole NaOH/2 mole
Al) x (40.0 grams NaOH/1 mole NaOH) = 87.6 grams NaOH
95.3 grams NaOH - 87.6 grams NaOH = 7.7 grams NaOH
d. (59.1 grams Al) x (1 mole Al/26.98 grams Al) x (3 mole H2/2 mole Al) x
(2.02 grams H2/1 mole H2) = 6.64 grams H2
6.
Fe3O4 + H2 → Fe + H2O
a. Fe3O4 + 4H2 → 3Fe + 4H2O
b. (.123 grams H2) x (1 mole H2/2.02 grams H2) x (4 mole H2O/4 mole H2)
x (18.02 grams H2O/1 mole H2O) = 1.10 grams H2O
Percent Yield = (0.744 grams/1.10 grams) x 100 = 67.64%