Question

How many g of fe can be produced by a reaction of 32.7 g co with excess fe2o3

Answer 1

Fe₂O₃ + 3CO $\to$ 2Fe + 3CO₂

$n=m/M$

$n(CO)= 32.7g/(28g/mol)=1.168mol$

From the chemical reaction equation:

$3n(Fe)=2n(CO)$

$n(Fe)=0.779 mol$

$m(Fe)=0.779mol*56g/mol=43.624g$

The answer is 43.624 grams