Answer:
55.85 grams of Fe is formed.
Step-by-step explanation:
Identify the reaction:
2Fe₂O₃ + 3C → 4Fe + 3CO₂
Identify the limiting reactant, previously determine the mol of each reactant
(mass / molar mass)
10 g / 12 g/m = 0.83 moles C
80 g / 159.7 g /m = 0.500 moles Fe₂O₃
2 moles of oxide need 3 moles of C, to react
0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol
I have 0.83 moles of C, so C is the excess.
The limiting is the oxide.
3 mol of C need 2 mol of oxide to react
0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)
Ratio is 2:4 (double)
If I have 0.5 moles of oxide, I will produce the double, in the reaction.
1 mol of Fe, will be produce so its mass is 55.85 g