# How many grams of sodium sulfide are formed if 1.50 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 94.0 % yield?

Feb 14, 2017

1.83g $N {a}_{2} S$

#### Explanation:

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.

2NaOH + H_2S → Na_2S + 2H_2O
So, it takes two moles of sodium hydroxide to react with one mole of hydrogen sulfide to produce one mole of sodium sulfide and two moles of water.

The actual amounts that we have are 1.5/34 = 0.044 moles of ${H}_{2} S$ and 2.0/40 = 0.05 moles of NaOH.

That means that NaOH is the “limiting reagent” - no more than 0.05/2 = 0.025 moles of ${H}_{2} S$ would be able to react theoretically.

Given the state 94% yield, this means that only 0.94 * 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 * 78 = 1.83g $N {a}_{2} S$.