Question

PLEASE HELP

One of the reactions for rusting iron is as follows: 4Fe + 3O2 → 2Fe2O3
(MM Fe: 55.85 g/mol; MM O2=32 g/mol; MM Fe2O3=159.70 g/mol)

If 2.80 moles of iron, Fe, rust completely, how many grams of iron (III) oxide are formed?
A. 0.0351 g
B. 313 g
C. 894 g
D. 224 g

Answer 1

It’s d. 224 g that’s the answer

Answer 2

D. 224 g

Step-by-step explanation:

Let's consider the rusting of iron reaction.

4 Fe + 3 O₂ → 2 Fe₂O₃

The molar ratio of Fe to Fe₂O₃ is 4:2. The moles of Fe₂O₃ produced from 2.80 moles of Fe are:

2.80 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) = 1.40 mol Fe₂O₃.

The molar mass of Fe₂O₃ is 159.70 g/mol. The mass of Fe₂O₃ is:

1.40 mol × (159.70 g / 1 mol) = 224 g