Question

a chemistry student obtained a yield of 3.80 g of H2O(l), at room temperature, from the reaction of 1.60 g H2(g) with 10.0 g O2 (g). what was her percentage yield of this reaction? (Hint: write a balanced reaction equation first and determine the limiting reagent?

Answer 1

2H₂+O₂=2H₂O

n(H₂)=m(H₂)/M(H₂)=1.6 g/2 g/mol = 0.8 mol

n(O₂)=m(O₂)/M(O₂)=10 g/32 g/mol = 0.3125 mol; for reacting with 0.3125 mol O₂ we need only 0.625 mol H₂; O₂ - limiting reagent.

in theory 0.3125 mol O₂ after reaction should give us 0.625 mol of H₂O - 0.625 * 18 (M H₂O)= 11.25 g. Then yield percentage = 3.80/11.25 *100% = 33.78%