1. 7.8 Limiting Reactants, Percent Yield
When you make a peanut butter sandwich, you need 2 slices of bread and 1 tablespoon of peanut butter. The reactant that runs out first is called the limiting reactant. Learning Goal Identify a limiting reactant and calculate the amount of product formed from the limiting reactant. Given the actual quantity of product, determine the percent yield for a reaction.

2. Limiting Reactant
A limiting reactant in a chemical reaction
is the substance that is used up first.
limits the amount of product that can form.
The reactant that does not completely react and is left over at the end of the reaction is called the excess reactant.

3. Calculating Moles of Product from a Limiting Reactant
In many reactions, there is a limiting reactant that determines the amount of product that can be formed.
Given a chemical reaction, from each reactant we can
calculate the amount of product possible when it is completely consumed.
determine the limiting reactant, the one that runs out first and produces the smaller amount of product.
Core Chemistry Skill Calculating the Quantity of Product from a Limiting Reactant

4. Guide to Calculating Moles of Product from a Limiting Reactant

5. Limiting Reactant, Moles of Product
If 3.00 moles of CO and 5.00 moles of H2 are the initial reactants, how many moles of methanol (CH3OH) can be produced and what is the limiting reactant? CO(g) + 2H2(g)  CH3OH(g) STEP 1 State the given and needed moles.
ANALYZE Given Need
THE PROBLEM moles of CO moles CH3OH produced moles of H2 limiting reactant
Equation
CO(g) + 2H2(g)  CH3OH(g)

6. Limiting Reactant, Moles of Product
STEP 2 Write a plan to convert the moles of each reactant to moles of product. moles of CO moles of CH3OH moles of H2 moles of CH3OH
Mole–mole factor
Mole–mole factor

7. Limiting Reactant, Moles of Product
STEP 3 Write the mole–mole factors from the equation.

8. Limiting Reactant, Moles of Product
STEP 4 Calculate the number of moles of product from each reactant, and select the smaller number of moles as the amount of product from the limiting reactant.
H2 is the limiting reactant, and 2.50 moles of CH3OH is produced.
Limiting
reactant
Smaller amount
of product

9. Limiting Reactant, Mass of Product
The quantities of reactants can also be given in grams. The calculations to identify the limiting reactant are
first, convert the grams of each reactant to moles of reactant using molar mass conversion factors;
second, use mole–mole factors to convert moles of reactant to moles of product; and
third, use molar mass to convert the moles of product to grams of product.
The reactant that produces the least amount of product is the limiting reactant.

10. Guide to Calculating the Grams of Product from a Limiting Reactant

11. Limiting Reactant, Mass of Product
Silicon carbide, SiC, is a ceramic material that tolerates extreme temperatures and is used as an abrasive and in the brake discs of sports cars. How many grams of CO are formed from 70.0 g of SiO2 and 50.0 grams of C? SiO2(s) + 3C(s)  SiC(s) + 2CO(g)
A ceramic brake disc in a sports car withstands temperatures of 1400 °C.

12. Limiting Reactant, Mass of Product
STEP 1 State the given and needed grams.
ANALYZE Given Need
THE PROBLEM g of SiO2 grams of CO produced g of C from limiting reactant
Equation
SiO2(s) + 3C(s)  SiC(s) + 2CO(g)

13. Limiting Reactant, Mass of Product
STEP 2 Write a plan to convert the grams of each reactant to grams of product.
grams moles moles grams
of SiO of SiO of CO of CO
grams moles moles grams
of H of H of CO of CO
Molar
mass
Mole–mole factor
Molar
mass
Molar
mass
Mole–mole factor
Molar
mass

14. Limiting Reactant, Mass of Product
STEP 3 Write the molar mass and mole–mole factors from the equation. Molar mass factors

15. Limiting Reactant, Mass of Product
STEP 4 Calculate the number of grams of product from each reactant, and select the smaller number of grams as the amount of product from the limiting reactant. C is the limiting reactant; 65.3 grams CO is produced.
Exact
Exact
Exact
Three SF
Three SF × × ×
Limiting
reactant
Smaller
amount of
product
Four SF
Exact
Four SF
Exact
Three SF
Four SF × × ×
Three SF
Four SF
Exact
Exact

16. Study Check
Given the following reaction, N2(g) + 3H2(g)  2NH3(g) calculate the amount of ammonia, NH3, that can be formed when 2.50 grams of nitrogen gas, N2, reacts with 2.00 grams of hydrogen gas, H2.

17. Solution STEP 1 State the given and needed grams. ANALYZE Given Need
THE PROBLEM g of N2 grams of NH3 produced g of H2 from limiting reactant
Equation
N2(g) + 3H2(g)  2NH3(g)

18. Solution
STEP 2 Write a plan to convert the grams of each reactant to grams of product.
grams moles moles grams
of N of N of NH of NH3
grams moles moles grams
of H of H of NH3 of NH3
Molar
mass
Mole–mole factor
Molar
mass
Molar
mass
Mole–mole factor
Molar
mass

19. Solution
STEP 3 Write the molar mass and mole–mole factors from the equation. Molar mass factors

20. Solution
STEP 3 Write the molar mass and mole–mole factors from the equation. Mole–mole factors

21. Solution
STEP 4 Calculate the number of grams of product from each reactant, and select the smaller number of grams as the amount of product from the limiting reactant. N2 is the limiting reactant; 3.04 grams of NH3 is produced.
Exact
Exact
Four SF
Three SF
Three SF × × ×
Limiting
reactant
Smaller amount of product
Four SF
Exact
Exact
Exact
Exact
Four SF
Three SF
Three SF × × ×
Three SF
Exact
Exact

22. Actual, Theoretical, and Percent Yield
When the reaction does not go to completion, or some of the reactant or product is lost, the amount of product produced may be less.
Theoretical yield is the maximum amount of product, which is calculated using the balanced equation.
Actual yield is the amount of product actually obtained.
Percent yield is the ratio of actual yield to theoretical yield.
Core Chemistry Skill Calculating Percent Yield

23. Guide to Calculations for Percent Yield

24. Study Check
On a space shuttle, LiOH is used to absorb exhaled CO2 from breathing air to form LiHCO3. What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO3? LiOH(s) + CO2(g)  LiHCO3(s)

25. Solution
On a space shuttle, LiOH is used to absorb exhaled CO2 from breathing air to form LiHCO3. What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO3? LiOH(s) + Co2(g) → LiHCO3(s) STEP 1 State the given and needed quantities.
ANALYZE Given Need
THE PROBLEM g of LiOH Percent yield LiHCO g of LiHCO Equation
LiOH(s) + CO2(g)  LiHCO3(s)

26. Solution
STEP 2 Write a plan to calculate theoretical yield and percent yield. grams moles moles grams of of of of LiOH LiOH LiHCO3 LiHCO3
Molar
mass
Mole–mole factor
Molar
mass

27. Solution
STEP 3 Write the molar mass and mole–mole factors from the equation. Molar mass factors

28. Solution
STEP 3 Write the molar mass and mole–mole factors from the equation. Mole-mole factors

29. Solution
STEP 4 Calculate the percent yield by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%. Calculation of percent yield:
Three SF
Exact
Exact
Four SF × × ×
Four SF
Exact
Exact
Three SF
Three SF × ×
Three SF