Answer:
80 g
Explanation:
We need to first find the limiting reactant and the excess one. To differentiate the two, we calculate how much of each reactant reacts with how much of the other to form Iron oxide.
According to the chemical equation, 4 moles of Iron react with 3 moles of Oxygen molecule to form 2 moles of ferric oxide. Since the molar mass of Fe, O2, and Fe2O3 are 56, 32, and 160 g/mole respectively, it means that 224g of Fe reacts with 96g of O2 to form 320g of Fe2O3.
[tex] \frac{57}{224} = \frac{x}{96} [/tex]
The above equation helps us to find how much of O2 (x) is needed to react fully with 57g of Fe which comes to be 24.43g which is higher than what we have (24g). Therefore O2 is the limiting reactant. To find the amount of Fe2O3 formed, we use the equation :
[tex] \frac{24}{96} = \frac{y}{320} [/tex]
y (the amount of Iron oxide formed) is 80g with 1g of Iron remaining still as 24g of O2 reacts completely with 56g of Fe.
