0.50 mol of Fe react with 0.60 mol of O2, how many moles of Fe2O3 will form?
4Fe + 3O2 ==> 2Fe2O3
0.50 mol Fe x (2 mol Fe2O3/4 mol Fe) = 0.50 mol Fe x (2/4) = ? mol Fe2O3.
To determine the number of moles of Fe2O3 that will form, we need to first determine the balanced chemical equation for the reaction between Fe and O2.
The balanced chemical equation for the reaction between Fe and O2 is:
4Fe + 3O2 -> 2Fe2O3
From the balanced equation, we can see that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.
Given that we have 0.50 mol of Fe and 0.60 mol of O2, we need to determine which reactant is limiting in order to calculate the amount of Fe2O3 formed.
To find the limiting reactant, we compare the ratios of the moles of reactants to the stoichiometric coefficients in the balanced equation.
For Fe:O2, the ratio is 4:3.
For the given amounts: 0.50 mol Fe : 0.60 mol O2
Let's calculate the number of moles of Fe2O3 formed using the limiting reactant:
Since the ratio of Fe:O2 is 4:3, we can use the number of moles of O2 to determine the number of moles of Fe required for complete reaction:
0.60 mol O2 * (4 mol Fe / 3 mol O2) = 0.80 mol Fe
Since 0.80 mol of Fe is greater than the amount of Fe given (0.50 mol), it means that Fe is present in excess and O2 is the limiting reactant.
Now, we can calculate the moles of Fe2O3 formed using the balanced equation:
From the balanced equation, we know that:
4 mol Fe reacts to produce 2 mol Fe2O3.
So, using the stoichiometric ratio:
0.60 mol O2 * (2 mol Fe2O3 / 3 mol O2) = 0.40 mol Fe2O3
Therefore, 0.40 moles of Fe2O3 will form when 0.50 mol of Fe reacts with 0.60 mol O2.