Chapter 10, Problem 7PE

Interpretation Introduction

Interpretation:

The number of milligrams of sodium hydroxide needed to produce $5.00\times {10}^2\text{mg}$ of sodium sulfate is to be calculated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction.

Answer to Problem 7PE

The number of milligrams of sodium hydroxide needed to produce $5.00\times {10}^2\text{mg}$ of sodium sulfate is $306\text{mg}$.

Explanation of Solution

The balanced equation for the reaction is given below.

$\text{S}\left(\text{s}\right)\text{+}\frac{\text{3}}{2}{\text{O}}_2\left(\text{g}\right)+\text{2NaOH(aq)}\to {\text{Na}}_2{\text{SO}}_4\text{(aq)}+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Therefore, $2$ millimoles of $\text{NaOH}$ produce $\text{1}$ millimole of ${\text{Na}}_2{\text{SO}}_4$.

Therefore millimole to millimole ratio is given below.

$\text{2}\text{mmol}\text{NaOH}=1\text{mmol}{\text{Na}}_2{\text{SO}}_4$

Therefore, two conversion factors from the millimole-to-millimole ratio are given below.

$\frac{\text{2}\text{mmol}\text{NaOH}}{1\text{mmol}{\text{Na}}_2{\text{SO}}_4}\text{and}\frac{1\text{mmol}{\text{Na}}_2{\text{SO}}_4}{\text{2}\text{mmol}\text{NaOH}}$

The conversion factor to obtain millimoles of $\text{NaOH}$ from ${\text{Na}}_2{\text{SO}}_4$ is given below.

$\frac{\text{2}\text{mmol}\text{NaOH}}{\text{1}\text{mmol}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}$

The molar mass of oxygen is $16.00\text{mg}{\text{mol}}^{-1}$.

The molar mass of sodium is $22.99\text{mg}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{mg}{\text{mol}}^{-1}$

Therefore, the molar mass of ${\text{Na}}_2{\text{SO}}_4$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 22.99\text{mg}{\text{mol}}^{-1}\right)+32.06\text{mg}{\text{mol}}^{-1}+\left(4\times \text{16}\text{.00}\text{mg}{\text{mol}}^{-1}\right)\\ & =& \text{45}\text{.98}\text{mg}{\text{mol}}^{-1}+32.06\text{mg}{\text{mol}}^{-1}+64.00\text{mg}{\text{mol}}^{-1}\\ & =& 142.04\text{mg}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain millimoles of ${\text{Na}}_2{\text{SO}}_4$ from milligrams of ${\text{Na}}_2{\text{SO}}_4$ is given below.

$\frac{\text{1}\text{mmol}{\text{Na}}_2{\text{SO}}_4}{\text{142}\text{.04}\text{mg}{\text{Na}}_2{\text{SO}}_4}$

The molar mass of oxygen is $16.00\text{mg}{\text{mol}}^{-1}$.

The molar mass of sodium is $22.99\text{mg}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.008\text{mg}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{NaOH}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 22.99\text{mg}{\text{mol}}^{-1}+16.00\text{mg}{\text{mol}}^{-1}+1.008\text{mg}{\text{mol}}^{-1}\\ & =& \text{40}\text{.00}\text{mg}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain milligrams of $\text{NaOH}$ from millimoles of $\text{NaOH}$ is given below.

$\frac{\text{40}\text{.00}\text{mg}\text{NaOH}}{\text{1}\text{mmol}\text{NaOH}}$

The formula to calculate the mass of $\text{NaOH}$ from ${\text{Na}}_2{\text{SO}}_4$ is given below.

$\text{Mass}\text{of}\text{NaOH}=\left(\begin{array}{l}\text{Actual}\text{mass}\text{of}{\text{Na}}_2{\text{SO}}_4\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain moles of }\text{NaOH}\\ \times \text{Conversion}\text{factor}\text{to obtain to}\text{obtain moles}\text{of}{\text{Na}}_2{\text{SO}}_4\\ \times \text{Conversion}\text{factor}\text{to obtain grams of}\text{NaOH}\end{array}\right)$…(1)

The percentage yield is calculated by the formula given below.

$\text{Percentage}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Ideal}\text{yield}}\times 100$…(2)

The percentage is $91.9\%$.

The actual yield is $5.00\times {10}^2\text{mg}$.

Substitute the value of actual yield and percentage in equation (2).

$\begin{array}{c}91.9\%=\frac{5.00\times {10}^2\text{mg}}{\text{Ideal}\text{yield}}\times 100\\ \text{Ideal}\text{yield}=\frac{5.00\times {10}^2\text{mg}}{91.9\%}\times 100\\ =544.1\text{mg}\end{array}$

Substitute the value of yield of ${\text{Na}}_2{\text{SO}}_4$ and conversion factors in the equation(1).

$\begin{array}{rll}\text{Mass}\text{of}\text{NaOH}& =& \left(\begin{array}{l}544.1\text{mg}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\times \frac{\text{2}\text{mmol}\text{NaOH}}{\text{1}\text{mmol}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \times \frac{\text{1}\text{mmol}{\text{Na}}_2{\text{SO}}_4}{\text{142}\text{.04}\text{mg}{\text{Na}}_2{\text{SO}}_4}\times \frac{\text{40}\text{.00}\text{mg}\text{NaOH}}{\text{1}\text{mmol}\text{NaOH}}\end{array}\right)\\ & =& \left(\begin{array}{c}544.1\text{mg}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\times 2\times \frac{\text{1}}{\text{142}\text{.04}\text{mg}{\text{Na}}_2{\text{SO}}_4}\\ \times \text{40}\text{.00}\text{mg}\text{NaOH}\end{array}\right)\\ & =& 306\text{mg}\end{array}$

Therefore, the number of milligrams of sodium hydroxide needed to produce $5.00\times {10}^2\text{mg}$ of sodium sulfate is $306\text{mg}$.

Conclusion

The number of milligrams of sodium hydroxide needed to produce $5.00\times {10}^2\text{mg}$ of sodium sulfate is $306\text{mg}$.