led concentrated HCI solution in excess to 45. nd collected the hydrogen I gas that was prod Mg(s)+2HCl(aq)arrow MgCl_(2)(aq)+ collected 2.01 grams of H_(2) . what was the per answer to the nearest : tenth of a Type your answer in the boxes square square square

To determine the percent yield of the reaction, we need to follow these steps:<br /><br />1. Calculate the theoretical yield of hydrogen gas (\(H_2\)).<br />2. Calculate the percent yield using the actual yield and the theoretical yield.<br /><br />First, let's write down the balanced chemical equation for the reaction:<br />\[ \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} \]<br /><br />From the balanced equation, we see that 1 mole of magnesium (\(Mg\)) produces 1 mole of hydrogen gas (\(H_2\)).<br /><br />Next, we need to find the molar mass of \(H_2\):<br />\[ \text{Molar mass of } H_2 = 2 \times 01 \, \text{g/mol} = 2.02 \, \text{g/mol} \]<br /><br />Now, let's calculate the theoretical yield of \(H_2\). Given that we have 45 grams of \(Mg\), we first need to convert this mass to moles:<br />\[ \text{Moles of } Mg = \frac{45 \, \text{g}}{24.31 \, \text{g/mol}} \approx 1.85 \, \text{moles} \]<br /><br />Since the reaction is 1:1, the moles of \(H_2\) produced will also be 1.85 moles. Now, we convert moles of \(H_2\) to grams:<br />\[ \text{Theoretical yield of } H_2 = 1.85 \, \text{moles} \times 2.02 \, \text{g/mol} \approx 3.72 \, \text{g} \]<br /><br />Finally, we calculate the percent yield using the actual yield (2.01 grams) and the theoretical yield (3.72 grams):<br />\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \]<br />\[ \text{Percent yield} = \left( \frac{2.01 \, \text{g}}{3.72 \, \text{g}} \right) \times 100\% \approx 53.8\% \]<br /><br />So, the percent yield of the reaction is approximately 53.8%.