Final answer:
In the decomposition of hydrogen peroxide, the theoretical yield of oxygen gas from 150.0 g of H₂O₂ is 70.56 g. Applying the percentage yield of 91.00%, the actual yield is 64.21 g of O₂.
Step-by-step explanation:
The question involves stoichiometry and the concept of percentage yield in a chemical reaction, specifically the decomposition of hydrogen peroxide (H₂O₂) to produce water (H₂O) and oxygen gas (O₂). To find the actual yield of oxygen, first, we need to calculate the theoretical yield, and then we apply the percentage yield.
The balanced chemical equation is: 2H₂O₂(aq) → 2H₂O(l) + O₂(g).
Molar mass of H₂O₂ is approximately 34.0147 g/mol. For 150.0 g of H₂O₂, moles of H₂O₂ = 150.0 g / 34.0147 g/mol ≈ 4.41 mol.
According to the balanced equation, 2 moles of H₂O₂ yields 1 mole of O₂, thus 4.41 mol of H₂O₂ would theoretically yield 2.205 mol of O₂.
Using the molar mass of O₂, which is approximately 32.00 g/mol, the theoretical yield in grams is 2.205 mol × 32.00 g/mol = 70.56 g.
Given a percentage yield of 91.00%, the actual yield of O₂ = 91.00% × 70.56 g
= 64.21 g (to two decimal places).