How many grams of hydrogen peroxide \( \left(\mathrm{H}_{2} \mathrm{O}_{2}\right) \) are needed to produce 25.0 g of oxygen? \( 2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \) 26.6 g 25.0 g 53.1 g 5.88 g Submit Request Answer

To solve for the mass of hydrogen peroxide needed to produce 25.0 g of oxygen, we need to start with the balanced chemical equation. According to the equation \(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\), we see that 2 moles of hydrogen peroxide yield 1 mole of oxygen. Calculating the molar mass of oxygen (\(O_2\)) gives us about 32.0 g/mol. Thus, to produce 25.0 g of \(O_2\): 1. Find moles of oxygen produced: \[ \text{Moles of } O_2 = \frac{25.0 \text{ g}}{32.0 \text{ g/mol}} \approx 0.78125 \text{ moles} \] 2. From the stoichiometry of the reaction, we need 2 moles of \(H_2O_2\) for every 1 mole of \(O_2\): \[ \text{Moles of } H_2O_2 = 2 \times 0.78125 \approx 1.5625 \text{ moles} \] 3. Now find the mass of \(H_2O_2\): - The molar mass of \(H_2O_2\) is approximately 34.0 g/mol. \[ \text{Mass of } H_2O_2 = 1.5625 \text{ moles} \times 34.0 \text{ g/mol} \approx 53.1 \text{ g} \] Therefore, to produce 25.0 g of oxygen, you would need about 53.1 g of hydrogen peroxide. The correct answer is **53.1 g**!