Given the equation for the decomposition of hydrogen peroxide, 2H_(2)O_(2)... O_(2)+2 H_(2)O, a student sets up a problem to convert from 63.4 grams of hydrogen peroxide to the mass of oxygen produced. Most of the setup is shown below. What is the final conversion factor? 63.4gH_(2)O_(2)times 1molH_(2)O_(2)times 1molO_(2)times 34.02gH_(2)O_(2) 2molH_(2)O_(2) 1molH_(2)O_(2) 1molO_(2) 32.00gO_(2) 1molO_(2) There isn't one - the problem is completely set up. 16.00gO_(2) 1molO_(2)

## Step 1: Analyze the given setup<br />### The problem involves converting 63.4 g of $H_2O_2$ to the mass of $O_2$. The equation for the decomposition is balanced as $2H_2O_2 \rightarrow O_2 + 2H_2O$. The student has already set up most of the conversion factors, but we need to verify if the final conversion factor is correct.<br /><br />---<br /><br />## Step 2: Identify the correct molar masses and stoichiometric ratios<br />### From the balanced equation:<br />- Molar mass of $H_2O_2 = 34.02 \, \text{g/mol}$.<br />- Stoichiometric ratio: $2 \, \text{mol} \, H_2O_2 \rightarrow 1 \, \text{mol} \, O_2$.<br />- Molar mass of $O_2 = 32.00 \, \text{g/mol}$.<br /><br />The correct sequence of conversions should be:<br />\[<br />63.4 \, \text{g} \, H_2O_2 \times \frac{1 \, \text{mol} \, H_2O_2}{34.02 \, \text{g} \, H_2O_2} \times \frac{1 \, \text{mol} \, O_2}{2 \, \text{mol} \, H_2O_2} \times \frac{32.00 \, \text{g} \, O_2}{1 \, \text{mol} \, O_2}.<br />\]<br /><br />---<br /><br />## Step 3: Verify the final conversion factor<br />### The final step in the calculation converts moles of $O_2$ to grams of $O_2$. This requires the molar mass of $O_2$, which is $32.00 \, \text{g/mol}$. Therefore, the correct final conversion factor is:<br />\[<br />\frac{32.00 \, \text{g} \, O_2}{1 \, \text{mol} \, O_2}.<br />\]<br /><br />---