Question

How many grams of sodium sulphide are formed if $1.50{\text{g}}$ of hydrogen sulphide is bubbled into a solution containing 2g of sodium hydroxide, assuming that the sodium sulphide is made in $94.0\%$ yield?

Answer 1

We are already provided with the equation. We need to first write the balanced equation. Then we shall calculate the moles of reactant given from the mass and use the concept of limiting reagents. The moles of limiting reagent shall be considered and used to find the mass of product formed.

Formula used: ${\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}}$

Complete step-by-step answer: The first step for any chemical reaction is to write out the chemical equation to determine the molar ratios of the reactants and products. Then we will use the molecular formulas and weights to calculate the desired answer.
The following equation is written as:
$2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O$
So, it takes two moles of sodium hydroxide to react with one mole of hydrogen sulphide to produce one mole of sodium sulphide and two moles of water.
The actual amounts that we have are $\dfrac{{1.5}}{{34}} = 0.44$ moles of ${H_2}S$ and $\dfrac{{2.0}}{{40}} = 0.05$ moles of NaOH.
That means that NaOH is the “limiting reagent” - no more than $\dfrac{{0.05}}{2} = 0.025$ moles of${H_2}S$ would be able to react theoretically.
Given the state $94.0\%$yield, this means that only $0.94 \times 0.025 = 0.0235$ moles of ${H_2}S$ will react to form $N{a_2}S$.
This will produce $0.0235$ moles of $N{a_2}S$.
The mass is found from the molecular weight again:
$0.0235 \times 78 = 1.83{\text{g}}$
So the mass of $N{a_2}S$is $1.83{\text{ g}}$.

Note: We already know that the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grams.
All the values should be put carefully and the equations should be solved neatly.