underline ( )H_(2)O_(2)arrow underline ( )H_(2)O+underline ( )O_(2) c. If you are given 25 g of H_(2)O_(2) how much Oxygen could theoretically be produced? d. If there were actually 9.41 g of Cl_(2) produced, what was the percent yield?

## Step 1: <br />The balanced chemical equation for the decomposition of hydrogen peroxide is given as:<br />### \(2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}\)<br /><br />## Step 2:<br />To find out how much oxygen could theoretically be produced from 25 g of \(H_{2}O_{2}\), we first need to calculate the molar mass of \(H_{2}O_{2}\). The molar mass of \(H_{2}O_{2}\) is approximately 34.0147 g/mol.<br /><br />## Step 3:<br />Next, we convert the given mass of \(H_{2}O_{2}\) to moles using the formula:<br />### \(n = \frac{m}{M}\)<br />where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass.<br /><br />## Step 4:<br />From the balanced chemical equation, we see that 2 moles of \(H_{2}O_{2}\) produce 1 mole of \(O_{2}\). Therefore, we can calculate the number of moles of \(O_{2}\) produced by multiplying the number of moles of \(H_{2}O_{2}\) by 1/2.<br /><br />## Step 5:<br />Finally, we convert the moles of \(O_{2}\) to grams using the molar mass of \(O_{2}\), which is approximately 31.998 g/mol.<br /><br />## Step 6:<br />To calculate the percent yield, we use the formula:<br />### \(\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\)<br />where the actual yield is the given amount of \(Cl_{2}\) produced, and the theoretical yield is the amount of \(O_{2}\) that could theoretically be produced from the given amount of \(H_{2}O_{2}\).