Question

Consider the following reaction: 2H2S (g) + 3O2 (g)  2SO2 (g) + 2H2O (g) If O2 was the excess reagent, 8.3 mol of H2S were consumed, and 137.1 g of water were collected after the reaction has gone to completion, what is the percent yield of the reaction? Show your work.

Answer 1

1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.  

8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced  

8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically  

% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield  

2. Calculate moles of each reactant.  

150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2  

32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2  

The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.  

So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.

Answer 2

The answer is: the percent yield of the reaction is 91.77%.

Balanced chemic reaction: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g).

n(H₂S) = 8.3 mol; amount of hydrogen sulfide.

m₁(H₂O) = 137.1 g; mass of water.

From chemical reaction: n(H₂S) : n(H₂O) = 2 : 2 (1 : 1).

n(H₂O) = 8.3 mol; amount of water.

m(H₂O) = 8.3 mol · 18 g/mol.

m(H₂O) = 149.4 g.

the percent yield = 137.1 g ÷ 149,4 g · 100%.

the percent yield = 91.77%.