Introduction:
The given chemical reaction is Fe3O4 + CO -> Fe + CO2. We need to determine the mass of Fe3O4 required to obtain 2 kg of Fe, considering the process to be 90% efficient.

Understanding the reaction:
The reaction shows that Fe3O4 reacts with CO to produce Fe and CO2. The molar ratio between Fe3O4 and Fe is 1:3, which means 1 mole of Fe3O4 is required to produce 3 moles of Fe.

Determining moles of Fe:
Given that we need to obtain 2 kg of Fe, we need to convert the mass of Fe to moles. The molar mass of Fe is approximately 55.85 g/mol. Therefore, the number of moles of Fe can be calculated as:
2 kg Fe * (1000 g/kg) / (55.85 g/mol) = 35.76 mol Fe

Calculating moles of Fe3O4:
Since the molar ratio between Fe3O4 and Fe is 1:3, the number of moles of Fe3O4 can be calculated by dividing the number of moles of Fe by 3:
35.76 mol Fe / 3 = 11.92 mol Fe3O4

Considering process efficiency:
The given process is stated to be 90% efficient. This means that only 90% of the Fe3O4 will actually react to produce Fe, and the rest will be wasted. Therefore, we need to account for this efficiency in our calculation.

Calculating mass of Fe3O4 required:
To determine the mass of Fe3O4 required, we need to reverse the molar ratio between Fe3O4 and Fe, taking into account the process efficiency.

11.92 mol Fe3O4 * (232.54 g/mol Fe3O4) / (0.9) = 3070.60 g Fe3O4

Therefore, approximately 3070.60 grams or 3.07 kg of Fe3O4 is required to obtain 2 kg of Fe.

Summary:
To obtain 2 kg of Fe with a 90% efficient process, approximately 3.07 kg of Fe3O4 is required. This calculation is based on the molar ratio between Fe3O4 and Fe, considering the process efficiency.