Question

What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe2O3 according to the equation: Fe2O3(s) + CO(g) → Fe(s) + CO2(g)?

A. 49.8 g Fe

B. 99.7 g Fe

C. 224 g Fe

D. 299 g Fe

Answer 1

B. 99.7 g Fe

Step-by-step explanation:

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Answer 2

Here we have to get the mass of the iron (Fe) produces in the reaction between ferric oxide (Fe₂O₃) and carbon monooxide (CO) to produce carbon dioxide and iron.

The amount of Fe produced in the reaction is B. 99.7g.

The balanced form of the reaction given is: Fe₂O₃ (s) + 3CO = 2Fe + 3CO₂.

Now as per the reaction 3 moles of CO produce 2 moles of Fe.

The molar mass of CO and Fe are 28.01g/mol and 55.845g/mol.

Thus (3×28.01) 84.03g of CO produces 111.69g of Fe.

Hence, 75.0g of CO produces
`(111.69)/(84.03)`×75.0 = 99.687g of Fe or 99.7g of Fe.

A. If in the reaction 49.8g Fe is produced in the reaction then
`(84.03)/(111.69)`×49.8 = 37.467g of CO should react. Thus the option is wrong.

C. To produce 224g of Fe
`(84.03)/(111.69)`×224 = 168.526g of CO will consumed. Thus it is wrong statement.

D. To generate 299g of Fe
`(84.03)/(111.69)`×299 = 224.952g of CO will consumed. Thus it is wrong statement.