How many grams of Fe are produced if the reaction begins with 3.0 grams of Al and 3.0 grams of FeO 2Al+3FeOarrow 1Al_(2)O_(3)+3Fe 103g 0.69 g 27.93 g 233g

## Step 1<br />The balanced chemical equation for the reaction is given as:<br />### \(2Al + 3FeO \rightarrow Al_{2}O_{3} + 3Fe\)<br />This equation tells us that 2 moles of Aluminum (Al) react with 3 moles of Iron(II) oxide (FeO) to produce 1 mole of Aluminum oxide (Al2O3) and 3 moles of Iron (Fe).<br /><br />## Step 2<br />The molar mass of Aluminum (Al) is approximately 26.98 g/mol, and the molar mass of Iron(II) oxide (FeO) is approximately 71.85 g/mol.<br /><br />## Step 3<br />We are given that 3.0 grams of Aluminum (Al) and 3.0 grams of Iron(II) oxide (FeO) are used in the reaction. We need to convert these amounts into moles using the formula:<br />### \(Moles = \frac{Mass}{Molar\,Mass}\)<br /><br />## Step 4<br />Using the above formula, we find that 3.0 grams of Aluminum (Al) is approximately 0.111 moles, and 3.0 grams of Iron(II) oxide (FeO) is approximately 0.042 moles.<br /><br />## Step 5<br />According to the balanced chemical equation, 2 moles of Aluminum (Al) react with 3 moles of Iron(II) oxide (FeO). Therefore, the ratio of Aluminum (Al) to Iron(II) oxide (FeO) is 2:3.<br /><br />## Step 6<br />We can use this ratio to find out how much Iron (Fe) will be produced. The ratio of Aluminum (Al) to Iron (Fe) is 2:3, so the amount of Iron (Fe) produced will be \(\frac{3}{2}\) times the amount of Aluminum (Al) used.<br /><br />## Step 7<br />Using the above ratio, we find that approximately 0.111 moles of Aluminum (Al) will produce approximately 0.111 moles of Iron (Fe).<br /><br />## Step 8<br />The molar mass of Iron (Fe) is approximately 55.85 g/mol. Therefore, 0.111 moles of Iron (Fe) will weigh approximately 0.111 moles * 55.85 g/mol = 6.19 grams.