The first step in solving this problem is to calculate the moles of Fe2O3 present:
moles of Fe2O3 = 21.4 g / 159.69 g/mol = 0.134 moles
Next, we need to determine which reactant is limiting, meaning it will be completely consumed in the reaction and determine the amount of product that can be formed. To do this, we can compare the moles of Fe2O3 to the moles of CO available for the reaction:
moles of CO = (3/1) * 0.134 moles = 0.402 moles
Since there are more moles of CO than are needed for the reaction, we know that Fe2O3 is the limiting reactant and all 0.134 moles will be used up. We can then calculate the moles of Fe that can be produced from the reaction:
moles of Fe = (2/1) * 0.134 moles = 0.268 moles
Finally, we use the molar mass of iron to convert the moles to grams:
mass of Fe = 0.268 moles * 55.85 g/mol = 15.0 grams
Therefore, the answer is (c) 15.0 grams.
How many grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3?
Fe2O3 + 3CO -----> 2Fe + 3CO2
a
0.150 grams
b
150 grams
c
15.0 grams
d
1.50 grams
1 answer
1 answer