Final answer:
To find out how many grams of Iron (III) oxide (Fe2O3) can be produced from 25.0 g of iron, we calculate the number of moles of iron, use the stoichiometric ratio from the balanced equation, and then convert the moles of Fe2O3 to grams to get the final yield of 35.68 g of Fe2O3.
Step-by-step explanation:
The question is asking how many grams of Iron (III) oxide (Fe2O3) can be formed from a given mass of iron when reacted with an excess of oxygen. To solve this, we need to use stoichiometry.
First, calculate the number of moles of iron using its molar mass (55.85 g/mol):
25.0 g Fe × (1 mol Fe / 55.85 g Fe) = 0.447 mol Fe
The balanced equation for the reaction is:
4 Fe + 3 O2 → 2 Fe2O3
This indicates that 4 moles of Fe produce 2 moles of Fe2O3. Using this stoichiometric ratio, we calculate the moles of Fe2O3 produced:
0.447 mol Fe × (1 mol Fe2O3 / 2 mol Fe) = 0.2235 mol Fe2O3
Now convert moles of Fe2O3 to grams using its molar mass (159.70 g/mol):
0.2235 mol Fe2O3 × (159.70 g Fe2O3 / 1 mol Fe2O3) = 35.68 g Fe2O3
So, 25.0 g of iron can produce 35.68 g of Iron (III) oxide.