Question

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 + O2--------- 2CO In the second step, iron(III) oxide and carbon monoxide react to form iron and carbon dioxide: Fe2O3 + 3CO ------- 2Fe + 3CO2Suppose the yield of the first step is and the yield of the second step is 79%. Calculate the mass of oxygen required to make 10g of Iron. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Answer 1

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