If a solution containing 118.08 g of mercury(II) nitrate is allowed to react completely with a...

Given:

• Amount of mercury (II) nitrate used in the reaction is 118.08 g.
• Amount of sodium sulfide solution is 16.642 g.

PART A

{eq}\begin{matrix} Reaction & Hg(NO_3)_2 & + & Na_2S & \longrightarrow & 2NaNO_3 & + & HgS \downarrow \\\\ Molar \ mass \ (M) & 324.7 \ g/mol & & 78 g/mol & & & & 232.66 \ g/mol \\ \\ Mass \ given \ (m) & 118.08 \ g & & 16.642 \ g & & & & \\\\ Moles \ reacting \ (\frac{m}{M})& \frac{118.08}{324.7} = 0.364 \ moles& & \frac{16.642}{78 } = 0.213 \ moles & & & & \boxed{0.213 \ moles} \\\\ \end{matrix} {/eq}

• Here we see that {eq}Hg(NO_3)_2 {/eq} is abundant while {eq}Na_2S {/eq} is the limiting reagent. So we perform the calculation with the amount of {eq}Na_2S {/eq}.

• Thus according to the reaction, we have 1 mole of {eq}Na_2S {/eq} react with 1 mole of {eq}Hg(NO_3)_2 {/eq} to give a precipitate of 1 mole of {eq}HgS {/eq}.

• Thus we have 0.213 mole of {eq}Na_2S {/eq} react with 0.213 mole of {eq}Hg(NO_3)_2 {/eq} to give a precipitate of 0.213 mole of {eq}HgS {/eq} which is equal to {eq}m = (0.213 \times 232.66) \ g = \boxed{49.55 \ g} {/eq}

PART B

Excess amount of {eq}Hg(NO_3)_2 {/eq} remains in the solution is {eq}n =( 0.364 - 0.213) \ moles = 0.151 \ moles {/eq} which is equivalent to {eq}m = (0.151 \times 324.7) \ g = \boxed {49.03 \ g} {/eq}