2) 400g of iron reacts with 6 moles of oxygen to form rust (iron III oxide)Determine the moles of rust formed and identify the limiting and excess reactant. Balanced Equation: __

## Step 1<br />The balanced chemical equation for the reaction of iron with oxygen to form rust (iron (III) oxide) is:<br />### \(4Fe + 3O_2 \rightarrow 2Fe_2O_3\)<br />This equation tells us that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of rust.<br /><br />## Step 2<br />We are given that 400g of iron is reacting with 6 moles of oxygen. We need to find out how many moles of iron are present in 400g. The molar mass of iron (Fe) is approximately 55.85g/mol. Therefore, the number of moles of iron is calculated as:<br />### \(400g / 55.85g/mol = 7.15 moles\)<br /><br />## Step 3<br />Now, we compare the number of moles of iron and oxygen. We can see that 7.15 moles of iron is more than the 4 moles of iron required by the balanced equation. However, we only have 6 moles of oxygen, which is less than the 3 moles of oxygen required by the balanced equation.<br /><br />## Step 4<br />Since we have less oxygen than required, oxygen is the limiting reactant. This means that the reaction will stop when all the oxygen is used up.<br /><br />## Step 5<br />To find the amount of rust formed, we use the stoichiometry of the reaction. According to the balanced equation, 3 moles of oxygen produce 2 moles of rust. Therefore, 6 moles of oxygen will produce:<br />### \(6 moles * (2 moles of rust / 3 moles of oxygen) = 4 moles of rust\)