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
Explanation
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Fe2o3-----4fe + 2o2
Mm of fe2o3= 320
Mm of fe in
d product. =224
:. 320-----224
remembering u were being given 70percent of fe2o3
70*80/100 = 56
320=224
56 = X so when u cross multiply u will have.
39.2
Option B
The iron ore contains 70% of Fe2O3
Mass of iron ore= 80kg
To obtain the mass of Fe2O3
70/100 X 80kg= 56kg
Since we already know the mass of Fe2O3 we can obtain the mass of Iron(Fe2);
Molar mass of Fe2O3= (2 x 56) + (16 x 3)= 160g
Molar mass of Fe2= (2 x 56)= 112g
To obtain the mass of Fe2;
Molar mass of Fe2/Molar mass of Fe2O3 x Mass of Fe2O3
112/160 x 56= 39.2kg
my school please clarify your answer to this question well biko,
its quite confusing.................
An iron ore is known to contain 70.0% Fe2O3. The mass of iron that can be obtain from 80kg of the ore is?
Ans: 39.2kg of Fe is obtain
(Above is the complete form of the question. You are to calculate the mass of iron obtain from 70% of 80kg of Fe2O3 ore)
