Practice Theoretical Yield in Chemistry

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

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Quick Recap

The maximum amount of product that could be formed in a chemical reaction, calculated from the stoichiometry of the balanced equation using the limiting reactant.

The perfect-world result — the most product you could possibly get if nothing is lost or wasted.

Showing a random 20 of 50 problems.

Example 1

medium
Compute the theoretical yield of NaCl (in g) when 2 mol Na reacts with 0.5 mol Cl2\text{Cl}_2 via 2Na+Cl2→2NaCl2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}. (MNaCl=58.5M_{NaCl}=58.5)

Example 2

hard
In Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2, theoretical yield (g) of Fe from 160 g Fe2O3\text{Fe}_2\text{O}_3 (M=160M=160) with excess CO. (MFe=56M_{Fe}=56)

Example 3

medium
In CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2, theoretical yield (g) of CaOCaO from 100 g CaCO3CaCO_3? (MCaCO3=100M_{CaCO_3}=100, MCaO=56M_{CaO}=56)

Example 4

easy
What is the theoretical yield?

Example 5

challenge
In C+O2→CO2C + O_2 \rightarrow CO_2 with 6 g C (M=12M=12) and 8 g O2O_2 (M=32M=32), find theoretical yield of CO2CO_2 (g). (M=44M=44)

Example 6

easy
In 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}, theoretical yield (mol) of H2O\text{H}_2\text{O} from 4 mol H2\text{H}_2 (excess O2\text{O}_2)?

Example 7

hard
Theoretical yield of a reaction is 25.0 g and actual yield is 18.0 g. Percent yield?

Example 8

easy
In N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, theoretical yield (g) of NH3\text{NH}_3 from 1 mol N2\text{N}_2 (excess H2\text{H}_2)? (MNH3=17M_{NH_3}=17)

Example 9

easy
In C+O2→CO2C + O_2 \rightarrow CO_2, what is the theoretical yield (mol) of CO2CO_2 from 2 mol C?

Example 10

medium
Calculate the theoretical yield of water when 8.08.0 g of hydrogen reacts with excess oxygen. (2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}; H = 1.0081.008, O = 16.00 g/mol16.00\,\text{g/mol})

Example 11

easy
Define theoretical yield and explain how it differs from actual yield.

Example 12

medium
In 2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O with 4 g H2H_2 (M=2M=2) and excess O2O_2, theoretical yield of water (g)? (M=18M=18)

Example 13

easy
In N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3, theoretical yield (mol) of NH3NH_3 from 2 mol N2N_2 (excess H2H_2)?

Example 14

easy
In 2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO, theoretical yield (g) of MgO from 1 mol Mg (excess O2O_2)? (MMgO=40M_{MgO}=40)

Example 15

easy
In 2KClO3→2KCl+3O22\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2, theoretical yield (mol) of O2\text{O}_2 from 2 mol KClO3\text{KClO}_3?

Example 16

hard
A student reacts 6.506.50 g of zinc with 20.020.0 mL of 3.00 M HCl3.00\,\text{M HCl}. The reaction is Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2. Determine the limiting reactant and the theoretical yield of ZnCl2\text{ZnCl}_2. (Zn = 65.3865.38, ZnCl2\text{ZnCl}_2 = 136.28 g/mol136.28\,\text{g/mol})

Example 17

easy
In C+O2→CO2\text{C} + \text{O}_2 \rightarrow \text{CO}_2, what is the theoretical yield (mol) of CO2\text{CO}_2 from 5 mol of C (excess O2\text{O}_2)?

Example 18

hard
In 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3, theoretical yield (g) of AlCl3\text{AlCl}_3 from 5.4 g Al (M=27M=27) and 21.3 g Cl2\text{Cl}_2 (M=71M=71)? (MAlCl3=133.5M_{AlCl_3}=133.5)

Example 19

easy
The theoretical yield is computed from the ____ reactant.

Example 20

hard
In CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}, theoretical yield (g) of CO2\text{CO}_2 from 8 g CH4\text{CH}_4 (M=16M=16, excess O2\text{O}_2)? (MCO2=44M_{CO_2}=44)
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