An iron ore is known to contain 70.0% Fe2O3. The obtain from 80kg of the ore is?
(Fe = 56, O = 16)
- 35.0 kg
-
39.2 kg
- 70.0 kg
- 78.4 kg
The correct answer is: B
Explanation
2Fe2O3 → 4Fe + 3O2320g 224g 96g
(70) / (100) x (80) / (1) 56g
if 0.32kg of Fe293 contains 0.224 kg
Then 56kg will contain (56 x 0.224) / (0.32) =39.2 kg