An iron ore is known to contain 70.0% Fe2O3. The obtain from 80kg of the ore is?(Fe = 56, O = 16)

Answer Details

The given iron ore contains 70.0% Fe2O3, which means that for every 100 grams of the ore, 70 grams is Fe2O3 and 30 grams is other substances. We are asked to determine the amount of Fe obtained from 80 kg of the ore. To solve this problem, we need to first calculate the amount of Fe2O3 in 80 kg of the ore. Since the ore contains 70.0% Fe2O3, the amount of Fe2O3 in 80 kg of the ore is: 70.0% × 80 kg = 0.70 × 80 kg = 56 kg Next, we need to calculate the amount of Fe in 56 kg of Fe2O3. The molecular weight of Fe2O3 is: 2(Fe) + 3(O) = 2(56 g/mol) + 3(16 g/mol) = 160 g/mol This means that 1 mole of Fe2O3 contains 2 moles of Fe. Therefore, the number of moles of Fe in 56 kg of Fe2O3 is: (56 kg) / (160 g/mol) × (2 mol Fe / 1 mol Fe2O3) = 0.7 mol Fe Finally, we can calculate the amount of Fe in 80 kg of the ore by multiplying the number of moles of Fe by the molar mass of Fe: 0.7 mol Fe × 56 g/mol = 39.2 kg Therefore, the amount of Fe obtained from 80 kg of the ore is 39.2 kg. Hence, the correct answer is option (B) 39.2 kg.