How many grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3?
Fe2O3 + 3CO -----> 2Fe + 3CO2
To find out how many grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3, we need to use stoichiometry and the balanced chemical equation.
1. Write down the balanced chemical equation:
Fe2O3 + 3CO → 2Fe + 3CO2
2. Determine the molar mass of Fe2O3:
Fe2O3 has a molar mass of 159.69 grams/mol (55.85 g/mol for each Fe x 2 + 16.00 g/mol for each O x 3)
3. Convert the given mass of Fe2O3 to moles:
21.4 g Fe2O3 * (1 mol Fe2O3 / 159.69 g Fe2O3) = 0.134 mol Fe2O3
4. Use stoichiometry to find the moles of Fe:
From the balanced equation, the stoichiometric ratio Fe2O3:Fe is 1:2.
So, the moles of Fe produced will be:
0.134 mol Fe2O3 * (2 mol Fe / 1 mol Fe2O3) = 0.268 mol Fe
5. Convert moles of Fe to grams:
The molar mass of Fe is 55.85 g/mol.
So, the mass of Fe produced will be:
0.268 mol Fe * (55.85 g Fe / 1 mol Fe) = 14.97 g Fe
Therefore, approximately 14.97 grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3.