Chapter 11, Problem 102A

Electrolysis Determine the theoretical and percent yield of hydrogen gas if 36.0 g ofwater undergoes electrolysis to produce hydrogen and oxygen and 3.80 g of hydrogen is collected.

Interpretation Introduction

Interpretation: Theoretical and Percent yield of hydrogen in the electrolysis of water is to be determined

Concept Introduction: Electrolysis of water is the process in which electricity is used to split water in hydrogen and oxygen gas.

Answer to Problem 102A

The theoretical yield of H₂ is 4 g.

Percent yield of H₂ is 95%.

Explanation of Solution

Step 1: Write balance chemical equation

  $\begin{array}{l}{\text{2H}}_2\text{O(l)}\to {\text{2H}}_2{\text{(g)+O}}_2\text{(g)}\end{array}$

  $\begin{array}{c}\text{36g}\text{3}\text{.8g}\end{array}$

Step 2: Convert the mass of reactant to moles

  $\begin{array}{c}Molarmassof{\text{H}}_2\text{O}=18g/mole\end{array}$

        $\begin{array}{c}Moles=\frac{Mass}{Molarmass}\end{array}$

     $\begin{array}{c}Molesof{\text{H}}_2\text{O}=\frac{36}{18}=2moles\end{array}$

Step 3: Convert moles of reactant to moles of product using a mole ratio.

2 mole of ${\text{H}}_2\text{O}$ produce = 2 moles of ${\text{H}}_{\text{2}}$

Step 4: Convert moles of product to mass (theoretical yield)

    $\begin{array}{c}\text{Molar}\text{}\text{mass}\text{of}{\text{H}}_2=\text{2 g/mole}\end{array}$

          $\begin{array}{c}\text{Mass}=\text{Molar}\text{mass}\times \text{Moles}\end{array}$

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2\text{(Theoretical}\text{yield)}=\text{2×2}\end{array}$

             $\begin{array}{c}=\text{ 4g}\end{array}$

Actual yield = 3.8 g (given)

$\begin{array}{c}\text{Percent}\text{Yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\text{×100\%}\end{array}$

      $\begin{array}{c}=\frac{3.8}{4}\times 100\%\end{array}$

      $\begin{array}{c}=95\%\end{array}$