Theoretical Yield Chemistry Example 4

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Example 4

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})

Solution

  1. 1
    Moles of Zn =6.5065.38=0.0994 mol= \frac{6.50}{65.38} = 0.0994\,\text{mol}. Moles of HCl =0.0200 L×3.00 M=0.0600 mol= 0.0200\,\text{L} \times 3.00\,\text{M} = 0.0600\,\text{mol}.
  2. 2
    The reaction requires 2 mol HCl per mol Zn. For 0.09940.0994 mol Zn, we need 0.1990.199 mol HCl, but only have 0.06000.0600 mol. HCl is limiting.
  3. 3
    From HCl: 0.0600 mol HCl×1 mol ZnCl22 mol HCl=0.0300 mol ZnCl20.0600\,\text{mol HCl} \times \frac{1\,\text{mol ZnCl}_2}{2\,\text{mol HCl}} = 0.0300\,\text{mol ZnCl}_2. Theoretical yield =0.0300×136.28=4.09 g= 0.0300 \times 136.28 = 4.09\,\text{g}.

Answer

HCl is limiting; theoretical yield=4.09 g of ZnCl2\text{HCl is limiting; theoretical yield} = 4.09\,\text{g of ZnCl}_2
When given amounts of both reactants, always check which is limiting before calculating the theoretical yield. The limiting reactant determines the maximum amount of product that can form.

About Theoretical Yield

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.

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More Theoretical Yield Examples

Example 1 easy

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

Example 2 medium

Calculate the theoretical yield of water when [formula] g of hydrogen reacts with excess oxygen. ([f

Example 3 medium

In the reaction [formula], calculate the theoretical yield of NaCl from [formula] g of sodium with e

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