Hydrogen peroxide (H2O2) undergoes the following decomposition reaction. 2H2O2 ⟶ 2H2O + O2 A scientist studying this reaction begins with 9.00 grams of H2O2 and recovers 4.00 grams of oxygen when the reaction is complete. What is the yield of the reaction? Round to the nearest 0.1% The reaction has a percent yield.

1. Calculate the molar mass of \( \text{H}_2\text{O}_2 \) and \( \text{O}_2 \).
- \( \text{Molecular weight of } \text{H}_2\text{O}_2 = 2(1.01) + 2(16.00) = 34.02 \) g/mol
- \( \text{Molecular weight of } \text{O}_2 = 2(16.00) = 32.00 \) g/mol

2. Calculate moles of \( \text{H}_2\text{O}_2 \) reacted.
\text{Moles of } \text{H}_2\text{O}_2 = \frac{9.00 \text{ g}}{34.02 \text{ g/mol}} = 0.2646 \text{ mol}

3. Based on the balanced equation, find moles of \( \text{O}_2 \) expected.
- From stoichiometry: \(2 \text{ mol } \text{H}_2\text{O}_2 \Rightarrow 1 \text{ mol } \text{O}_2\)
- Expected moles of \( \text{O}_2 = 0.2646 \div 2 = 0.1323 \text{ mol} \)

4. Convert moles of \( \text{O}_2 \) to grams.
\text{Grams of expected } \text{O}_2 = 0.1323 \text{ mol} \times 32.00 \text{ g/mol} = 4.2336 \text{ g}

5. Calculate percent yield.
\text{Percent yield} = \left(\frac{4.00 \text{ g}}{4.2336 \text{ g}}\right) \times 100\% = 94.441\%

6. Round to the nearest 0.1\%
- The reaction has a 94.1\% yield.