Question

Mercury and its compounds have uses from fillings for teeth (as a mixture with silver, copper, and tin) to the production of chlorine. Because of their toxicity, however, soluble mercury compounds, such as mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.145 L of 0.019 M mercury(II) nitrate reacts with 0.236 L of 0.048 M sodiium sulfide. What mass of mercury(II) sulfide is formed? Hint: Write a balanced chemical equation. Enter to 4 decimal places.

Answer 1

Answer:

0.6410g of HgS (mercury (II) sulfide) are formed

Explanation:

First you should write the balanced chemical equation, so we have:

$Na_{2}S+Hg(NO_{3})_{2}=HgS+_{2}NaNO_{3}$

Where:

$Na_{2}S$ is the formula for the sodium sulfide

$Hg(NO_{3})_{2}$ is the formula for the mercury (II) nitrate

$HgS$ is the formula for the mercury (II) sulfide

and  

$NaNO_{3}$ is the formula for the sodium nitrate

Then you should calculate the amount of each substance in each solution, so:

- For the $Na_{2}S$:

$Molarity=\frac{n_{Na_{2}S}}{Lofsolution}$

$n_{Na_{2}S}=0.048M*0.236L$

$n_{Na_{2}S}=0.010856$ moles of $Na_{2}S$

- For the $Hg(NO_{3})_{2}$:

$Molarity=\frac{n_{Hg(NO_{3})_{2}}}{Lofsolution}$

$n_{Hg(NO_{3})_{2}}=0.019M*0.145L$

$n_{Hg(NO_{3})_{2}}=0.002755$ moles of $Na_{2}S$

As the quantity of $Hg(NO_{3})_{2}$ is smaller than the quantity of $Na_{2}S$. The $Hg(NO_{3})_{2}$ is the limiting reagent and you should work with this quantity, so we have:

$0.002755molesHg(NO_{3})_{2}*\frac{1molHgS}{1molHg(NO_{3})_{2}}=0.002755$ moles of HgS

And as the molar mass of the HgS is $232.66\frac{g}{mol}$ you can calculate the mass of HgS that is produced:

$0.002755molesHgS*\frac{232.66gHgS}{1molHgS}=0.6410g$ HgS