Question

What mass of hydrogen sulfide will be needed to precipitate $7.5\mathrm{g}$ of copper sulfide from a copper sulfate solution?

Answer 1

Step 1: Given information

• Mass of copper sulfide $\left(\mathrm{CuS}\right)=7.5\mathrm{g}$

Step 2: Reaction involved

• When hydrogen sulfide $\left({\mathrm{H}}_2\mathrm{S}\right)$ gas is passed through a blue copper sulfate $\left({\mathrm{CuSO}}_4\right)$ solution, a black copper sulfide $\left(\mathrm{CuS}\right)$ precipitate is formed, and the resulting sulfuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ remains in the solution.
• The reaction is as follows:

${\mathrm{CuSO}}_4+{\mathrm{H}}_2\mathrm{S}\to \mathrm{CuS}\downarrow +{\mathrm{H}}_2{\mathrm{SO}}_4$

Step 3: Balanced chemical reaction

• The balanced chemical reaction is as follows:

${\mathrm{CuSO}}_4+{\mathrm{H}}_2\mathrm{S}\to \mathrm{CuS}\downarrow +{\mathrm{H}}_2{\mathrm{SO}}_4$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{H}}_2\mathrm{S}& =& 1\mathrm{g}\times 2+32\mathrm{g}\\ & =& 34\mathrm{g}\end{array}$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}\mathrm{CuS}& =& 63.5\mathrm{g}+32\mathrm{g}\\ & =& 95.5\mathrm{g}\end{array}$

Step 4: Mass of hydrogen sulfide

• $95.5\mathrm{g}$ of copper sulfide require $34\mathrm{g}$ hydrogen sulfide for precipitation.
• To precipitate $7.5\mathrm{g}$ of copper sulfide require $\frac{34\mathrm{g}}{95.5\mathrm{g}}\times 7.5\mathrm{g}=2.67\mathrm{g}$ of hydrogen sulfide

Therefore, $2.67\mathrm{g}$ of hydrogen sulfide is needed to precipitate $7.5\mathrm{g}$ of copper sulfide from a copper sulfate solution.