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?
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.
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
That means that NaOH is the “limiting reagent” - no more than 0.05/2 = 0.025 moles of
Given the state 94% yield, this means that only 0.94 * 0.025 = 0.0235 moles of
This will produce 0.0235 moles of
The mass is found from the molecular weight again:
0.0235 * 78 = 1.83g