Chapter ST, Problem 27PQ

Interpretation Introduction

Interpretation:

Moles of ${\text{N}}_2{\text{H}}_4$ produced from reaction of ${\text{NH}}_3$ with ${\text{OCl}}^{-}$ has to be determined.

Concept Introduction:

Ratio of actual yield by theoretical yield multiplied by 100 gives percent yield. Percent yield equal to $100\%$ implies same actual and theoretical yield. Usually, percent yield is obtained below $100\%$ because actual yield lower in magnitude than theoretical value due to incomplete reactions and loss of sample during recovery.

The equation for percent yield is as follows:

  $\text{percent yield }=\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right)\times 100\%$

Here

Actual yield is quantity of product obtained from a chemical reaction

Theoretical yield is quantity of product obtained from balanced equation.

Units for both actual and theoretical yield need to be the same (moles or grams).

Answer to Problem 27PQ

Option (A) is the correct one.

Explanation of Solution

Reason for correct option:

The reaction of ${\text{NH}}_3$ with ${\text{OCl}}^{-}$ is as follows:

  ${\text{2NH}}_3+{\text{OCl}}^{-}\to {\text{N}}_2{\text{H}}_4+{\text{Cl}}^{-}+{\text{H}}_2\text{O}$

$2\text{mol}$ of ${\text{NH}}_3$ produces $1\text{mol}$ of ${\text{N}}_2{\text{H}}_4$ so moles of ${\text{N}}_2{\text{H}}_4$ produced from $5.85\text{mol}$ of ${\text{NH}}_3$ can be calculated as follows:

  $\begin{array}{c}\text{moles}\text{of}{\text{N}}_2{\text{H}}_4=\left(\frac{1\text{mol}}{\text{2 mol}}\right)\left(5.85\text{mol}\right)\\ =\left(0.5\right)\left(5.85\text{mol}\right)\\ =2.925\text{mol}\end{array}$

Formula to determine the weight of ${\text{N}}_2{\text{H}}_4$ is as follows:

  $\text{mass}\text{of}{\text{N}}_2{\text{H}}_4=\left(\text{moles}\text{of}{\text{N}}_2{\text{H}}_4\right)\left(\text{molecular}\text{weight}\text{of}{\text{N}}_2{\text{H}}_4\right)$        (1)

Substitute $2.925\text{mol}$ for moles of ${\text{N}}_2{\text{H}}_4$ and $32.0452\text{g/mol}$ for molecular weight of ${\text{N}}_2{\text{H}}_4$ in equation (1) to determine mass of ${\text{N}}_2{\text{H}}_4$.

  $\begin{array}{c}\text{mass}\text{of}{\text{N}}_2{\text{H}}_4=\left(2.925\text{mol}\right)\left(32.0452\text{g/mol}\right)\\ =93.7322\text{g}\end{array}$

The equation for percent yield of ${\text{N}}_2{\text{H}}_4$ is as follows:

  $\text{percent yield}=\left(\frac{{\text{actual yield of N}}_2{\text{H}}_4}{\text{theoretical yield}{\text{of N}}_2{\text{H}}_4}\right)$        (2)

Rearrange equation (2) to calculate the actual yield of ${\text{N}}_2{\text{H}}_4$.

  ${\text{actual yield of N}}_2{\text{H}}_4=\left(\text{percent yield}\right)\left(\text{theoretical yield}{\text{of N}}_2{\text{H}}_4\right)$        (3)

Substitute $93.7322\text{g}$ for theoretical yield of ${\text{N}}_2{\text{H}}_4$ and $78.2\%$ for percent yield in equation (3) to determine actual yield of ${\text{N}}_2{\text{H}}_4$.

  $\begin{array}{c}{\text{actual yield of N}}_2{\text{H}}_4=\left(\text{78}\text{.2}\text{\%}\right)\left(93.7322\text{g}\right)\\ =73.29\text{g}\end{array}$

The formula to determine moles of ${\text{N}}_2{\text{H}}_4$ is as follows:

  $\text{moles}\text{of}{\text{N}}_2{\text{H}}_4=\frac{\text{mass}\text{of}{\text{N}}_2{\text{H}}_4}{\text{molecular}\text{weight}\text{of}{\text{N}}_2{\text{H}}_4}$        (4)

Substitute $73.29\text{g}$ for mass of ${\text{N}}_2{\text{H}}_4$ and $32.0452\text{g/mol}$ for molecular weight of ${\text{N}}_2{\text{H}}_4$ in equation (4) to determine moles of ${\text{N}}_2{\text{H}}_4$.

  $\begin{array}{c}\text{moles}\text{of}{\text{N}}_2{\text{H}}_4=\frac{73.29\text{g}}{32.0452\text{g/mol}}\\ =2.29\text{mol}\end{array}$

Moles of ${\text{N}}_2{\text{H}}_4$ produced from reaction is $2.29\text{mol}$. Hence, correct option is (A).

Reason for incorrect option:

Moles of ${\text{N}}_2{\text{H}}_4$ produced from reaction is $2.29\text{mol}$. Therefore, choice (B), (C) and (D) is incorrect.

Conclusion

Moles of ${\text{N}}_2{\text{H}}_4$ produced from reaction of ${\text{NH}}_3$ with ${\text{OCl}}^{-}$ is $2.29\text{mol}$. Hence, correct option is (A).