4Fe + 3O2 = 2Fe2O3
n=m/M
n(Fe)= 60g/55.85g/mol = 1.074 mol
n(O2)= 26g / 32.00g/mol = 0.8125 mol
Find limiting reactant. Compare:
n(Fe)/4 and n(O2)/3
1.074/4 and 0.8125/3
0.2685<0.2708
So, Fe is a limiting reactant. Use amount of substance of Fe to find amount of substance of Fe2O3.
1.074/4 = n(Fe2O3)/2
n(Fe2O3)= 0.537 mol
m=M*n
M(Fe2O3) = 55.85*2+16.00*3= 159.7 g/mol
m(Fe2O3) = 0.537 mol *159.7g/mol = 85.76 g = 86g.
Answer: 86 g
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