Chapter 3, Problem 3.139QE

(a)

Interpretation Introduction

Interpretation:

The percent yield of $\text{12}\text{.9}\text{g}{\text{NaNO}}_{\text{3}}$ formed from the reaction between $\text{23}\text{.1}\text{g}\text{NaOH}$ and $\text{21}\text{.2}\text{g}{\text{HNO}}_{\text{3}}$ has to be determined.

Explanation of Solution

The balanced equation for the reaction between $\text{NaOH}$ and ${\text{HNO}}_{\text{3}}$ to yield ${\text{NaNO}}_{\text{3}}$ is:

    ${\text{NaOH+HNO}}_{\text{3}}\stackrel{}{\to }{\text{NaNO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$.

Number of moles of each reactant has to be determined:

    $\text{Amount NaOH = 23}\text{.1g NaOH × }\frac{\text{1mol NaOH}}{\text{39}\text{.997}}\text{= 0}\text{.5775 moles}$

    ${\text{Amount HNO}}_3\text{ = 21}{\text{.2g HNO}}_3\text{ × }\frac{{\text{1mol HNO}}_3}{\text{63}\text{.01}}\text{= 0}\text{.336 moles}$.

Then calculate the equivalent amount of ${\text{NaNO}}_{\text{3}}$ produced from the moles of each reactant.  The reactant that yields smaller number of ${\text{NaNO}}_{\text{3}}$ is the limiting reactant.

${\text{Amount NaNO}}_{\text{3}}\text{based on NaOH = 0}\text{.5775 mol NaOH× }\left(\frac{{\text{1mol NaNO}}_{\text{3}}}{\text{1mol NaOH}}\right)\text{=0}{\text{.5775 mol NaNO}}_{\text{3}}$

${\text{Amount NaNO}}_{\text{3}}{\text{based on HNO}}_3\text{ = 0}{\text{.336 mol HNO}}_3\text{× }\left(\frac{{\text{1mol NaNO}}_{\text{3}}}{{\text{1mol HNO}}_3}\right)\text{=0}{\text{.336 mol NaNO}}_{\text{3}}$.

${\text{HNO}}_{\text{3}}$ is the limiting reactant, less amount of ${\text{NaNO}}_{\text{3}}$ is produced from it.

Use the number of moles of the limiting reactant to calculate the yield:

    ${\text{Mass NaNO}}_{\text{3}}\text{= 0}{\text{.336mol NaNO}}_{\text{3}}\left(\frac{\text{84}{\text{.9947g NaNO}}_{\text{3}}}{{\text{1mol NaNO}}_{\text{3}}}\right)\text{ = 28}{\text{.558g NaNO}}_{\text{3}}$.

The percent yield is the actual yield of the reaction divided by the theoretical yield, times $100\%$.

    $\text{Percent yield = }\frac{\text{12}{\text{.9g NaNO}}_{\text{3}}\text{(actual)}}{\text{28}{\text{.559g NaNO}}_{\text{3}}\text{(theoretical)}}\text{×100\% = 45}\text{.17\%}$

(b)

Interpretation Introduction

Interpretation:

A $\text{12}\text{.9}\text{g}{\text{NaNO}}_{\text{3}}$ is formed from the reaction between $\text{23}\text{.1}\text{g}\text{NaOH}$ and $\text{21}\text{.2}\text{g}{\text{HNO}}_{\text{3}}$.  The reactant that is present in excess has to be identified and the mass of that reactant that remains after the reaction has so be calculated.

Explanation of Solution

The balanced equation for the reaction between $\text{NaOH}$ and ${\text{HNO}}_{\text{3}}$ to yield ${\text{NaNO}}_{\text{3}}$ is:

    ${\text{NaOH+HNO}}_{\text{3}}\stackrel{}{\to }{\text{NaNO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$.

Number of moles of each reactant has to be determined:

    $\text{Amount NaOH = 23}\text{.1g NaOH × }\frac{\text{1mol NaOH}}{\text{39}\text{.997}}\text{= 0}\text{.5775 moles}$

    ${\text{Amount HNO}}_3\text{ = 21}{\text{.2g HNO}}_3\text{ × }\frac{{\text{1mol HNO}}_3}{\text{63}\text{.01}}\text{= 0}\text{.336 moles}$.

The reactant that is present in excess is sodium hydroxide.

  $\begin{array}{l}{\text{Amount NaNO}}_{\text{3}}\text{ produced =0}{\text{.336 mol NaNO}}_{\text{3}}\\ \text{Number of moles of NaOH reacted = 0}\text{.336mol}\\ \text{Mass of NaOH =0}\text{.336mol×39}\text{.997g/mol=13}\text{.4g}\end{array}$.

Mass of sodium hydroxide remained is $\text{23}\text{.1g-13}\text{.4g = 9}\text{.66g}$.