2. underline ( )Al+underline ( )Fe_(2)O_(3)arrow underline ( )Al_(2)O_(3)+underline ( )Fe a.If you have 15 grams of Al, what mass of Al_(2)O_(3) could theoretically be produced? b. If there was actually 19.27 g of Al_(2)O_(3) produced, what was the percent yield?

## Step 1<br />The balanced chemical equation for the reaction is:<br />### \(2Al + Fe_{2}O_{3} \rightarrow Al_{2}O_{3} + 2Fe\)<br />This equation tells us that 2 moles of Aluminum (Al) react with 1 mole of Iron(III) oxide (Fe2O3) to produce 1 mole of Aluminum oxide (Al2O3) and 2 moles of Iron (Fe).<br /><br />## Step 2<br />To find out how much Al2O3 can be produced from 15 grams of Al, we first need to convert the mass to moles. The molar mass of Al is approximately 26.98 g/mol. So, the number of moles of Al is:<br />### \(15g / 26.98g 0.56 moles\)<br /><br />## Step 3<br />According to the balanced chemical equation, 2 moles of Al produce 1 mole of Al2O3. Therefore, 0.56 moles of Al will produce28 moles of Al2O3.<br /><br />## Step 4<br />The molar mass of Al2O3 is approximately 101.96 g/mol. Therefore, the mass of Al2O3 produced is:<br />### \(0.28 moles * 101.96 g/mol = 28.55g\)<br /><br />## Step 5<br />The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. In this case, the actual yield is 19.27g and the theoretical yield is 28.55g. So, the percent yield is:<br />### \((19.27g / 28.55g) * 100 = 67.8%\)