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

## Question:

If a solution containing 118.08 g of mercury(II) nitrate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed?

## Limiting Reactant:

Stoichiometry deals with the calculation of the amounts of reactants and the products used as per the balanced chemical equation.

Now, a definite amount of a particular substance reacts with a fixed amount of another substance to get the products. Now if the amount of one reactant is in excess, the fraction which is in excess remains as it is after the reaction. The reactant with a lesser amount is the limiting one and is fully consumed.

## Answer and Explanation: 1

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}

#### Learn more about this topic:

from

Chapter 9 / Lesson 5Learn how to find the limiting and excess reactants in a chemical reaction. See example problems that calculate the limiting and excess reactants.