Question

1. Problem Set 4 Chem 141 When hydrogen sulfide (H.S. MM - 34,08 g/mol) gas is bubbled into a solution of sodium hydroxide (NaOH. 40.00 moll, sodium sulfide (Na S. 78.04 g/ mol) and water (18.02 g/mol) are produced according to the balanced chemical equation shown below? HS + 2NaOH-.. Nas + 2 HO (a) Assuming the reaction goes to completion, how many grams of sodium sulfide are formed if 2.50g of hydrogen sulfide is bubbled into a solution containing 1.85g of NaOH? (20 pts) (b) Which reactant and how much of it remains after the reaction has been completed? (15 pts) (@) If only 0.400g of sodium sulfide was recovered, what is the percent yield of this reaction (5 pts)

Answer 1

1. a) Number of moles of H₂S=Given mass/Molar mass=2.50 g/34.08 g/mol=0.073 mol

Number of moles of NaOH=Given mass/molar mass

=1.85 g/40.00 g/mol=0.046 mol

As per the balanced chemical equation

1 mol H₂S reacts with 2 mol NaOH

So 0.073 mol H₂S reacts with 2x0.073 mol=0.146 mol NaOH

But we have only 0.046 mol NaOH, so NaOH is the limiting reagent

Now 2 mol NaOH produces 1 mol Na₂S

1 mol NaOH produces 1/2 mol Na₂S

So 0.046 mol NaOH produces (1/2)x0.046 mol=0.023 mol Na₂S

Mass of Na₂S formed=Number of moles x molar mass of Na₂S=0.023 mol x 78.04 g/mol=1.79 g

So 1.79 g sodium sulphide will be formed

b)As NaOH is the limiting reagent, H₂S is in excess

As per the balanced chemical equation

2 mol NaOH reacts with 1 mol H₂S

1 mol NaOH reacts with 1/2 mol H₂S

0.046 mol NaOH reacts with (1/2)x0.046 mol=0.023 mol H₂S

Mass of H₂S that reacts with given amount of NaOH=Number of moles x molar mass of H₂S

=0.023 mol x 34.08 g/mol=0.78 g

So remaining amount of H₂S=2.50 g -0.78 g=1.72 g

so 1.72 g H₂S remains after the reaction

c. Percent yield=(Actual yield/Theoretical yield)x100

=(0.400 g/1.79 g)x100=22.35 %