Final answer:
To find the percent yield of H₂O when 14.5L H2(g) reacted with excess O₂(g) to yield 9.52g H₂O, the theoretical yield of H₂O was first calculated to be 11.66 g. The percent yield was then determined to be about 81.65% by comparing the actual yield to the theoretical yield.
Step-by-step explanation:
The student asked about the percent yield of H₂O when 14.5L H₂(g) reacted with excess O₂(g) at STP to yield 9.52g H₂O. To calculate the percent yield, we must first determine the theoretical yield of H₂O that should have been produced from 14.5L of H₂(g).
Using the balanced chemical equation:
2H₂(g) + O₂(g) → 2H₂O(g),
and knowing that at STP 1 mol of gas occupies 22.4L, we calculate the moles of H₂ reacted (14.5 L / 22.4 L/mol = 0.64732 mol H₂), which should produce an equal amount of moles of H₂O, since they are in a 1:1 ratio in the chemical equation.
The molar mass of H₂O is 18.02 g/mol, so the theoretical yield is 0.64732 mol × 18.02 g/mol = 11.66 g of H₂O. The percent yield is then calculated using the actual yield (9.52 g) divided by the theoretical yield (11.66 g) and multiplied by 100, which gives us approximately 81.65%.