Final answer:
The actual yield of oxygen from the decomposition of 100. g of H₂O₂ with a 93.0% yield is 43.7 g, which corresponds to option a.
Step-by-step explanation:
The question asks about the actual yield of oxygen gas when decomposing hydrogen peroxide (H₂O₂) if the percentage yield is 93.0%.
Given that we start with 100. g of H₂O₂, we first need to find out how many moles that is. Hydrogen peroxide has a molar mass of approximately 34.01 g/mol, so we can calculate the number of moles of H₂O₂ as 100 g / 34.01 g/mol, which is roughly 2.94 moles.
According to the balanced equation for the decomposition of hydrogen peroxide, 2 moles of H₂O₂ produce 1 mole of O₂. So, 2.94 moles of H₂O₂ would theoretically produce 1.47 moles of O₂.
Next, we find the theoretical yield in grams by multiplying the moles of O₂ by its molar mass, which is approximately 32.00 g/mol.
Therefore, 1.47 moles × 32.00 g/mol gives us a theoretical yield of 47.04 g of O₂.
Applying the percentage yield, we calculate the actual yield: 47.04 g × 0.930 (93.0%) = 43.7 g.
Among the answer options, the closest to 43.7 g is option a, which is 43.7 g.