Answer:
The mass of O2 that will produce 10 g of Iron, Fe is 6.0 g
Note: The question is missing some details. The complete question is given below:
Blast furnaces extra pure iron from the iron (III) oxide in iron ore in a two step sequence. In the first step, carbon and oxygen react to form carbon monoxide 2C(s)+O2(e) -2CO(« In the second step, iron(III) oxide and carbon monoxide react to form iron and carbon dioxide Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3CO2(g) Suppose the yield of the first step is 90.% and the yield of the second step is 79.%. Calculate the mass of oxygen required to make 10 g of iron. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.
Explanation:
Equation for the two reactions is given below:
Step 1: 2C(s) + O2(g) → 2CO(g)
Step 2: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
In the second step 3 moles of CO reacts to produce 2 moles of Fe
Molar mass of CO = 28 g/mol, molar mass of Fe = 56 g/mol
Therefore, 84 g (3 × 28) of CO will produce 112 g (2 × 56) of Fe
10 g of Fe will be produced by 84/112 × 10 g of CO = 7.5 g of CO
However, the actual yield is 79%, therefore, 7.5 g of CO will produce 0.79 × 10 g of Fe = 7.9 g of Fe
Mass of CO that will produce an actual yield of 10 g of Fe = 7.5/7.9 × 10 = 9.50 g of CO
From the first step:
1 mole of O2 reacts to produce 2 moles of CO (molar mass of O2 = 32 g)
32 g of O2 produces 56 g (2 × 28 g) of CO
Mass of O2 that will produce 9.50 g of CO = 32/56 × 9.50 g = 5.43g
However, since the actual yield is 90%, therefore, mass of CO produced = 0.9 × 9.50 = 8.55 g of CO
Mass of O2 that will produce 9.50 g of CO = 5.43/8.55 × 9.50 g = 6.03 g
Therefore, mass of O2 that will produce 10 g of Iron, Fe is 6.0 g
