Fe2O3+ 3CO=2Fe+3Co2, suppose that 1.64kg of Fe are obtained from a 2.62kg sample of Fe2O3 (ore). assuming that the reaction goes to completion, what is the percent purity of the Fe2O3 in the original sample.

To determine the percent purity of Fe2O3 in the original sample, we need to compare the amount of Fe obtained to the maximum amount that could be obtained if all the Fe2O3 reacted.

1. Calculate the molar mass of Fe2O3:
- Molar mass of Fe = 55.845 g/mol
- Molar mass of O = 16.00 g/mol (3 O atoms in Fe2O3)
- Molar mass of Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol

2. Convert the mass of Fe obtained to moles:
- Mass of Fe obtained = 1.64 kg = 1640 g
- Moles of Fe = mass of Fe obtained / molar mass of Fe = 1640 g / 55.845 g/mol = 29.35 mol

3. Use the stoichiometry of the balanced equation to find the moles of Fe2O3 that reacted:
- From the balanced equation: 1 mol Fe2O3 : 2 mol Fe
- Moles of Fe2O3 reacted = 29.35 mol Fe * (1 mol Fe2O3 / 2 mol Fe) = 14.68 mol Fe2O3

4. Convert the moles of Fe2O3 reacted to mass:
- Mass of Fe2O3 reacted = moles of Fe2O3 reacted * molar mass of Fe2O3 = 14.68 mol * 159.69 g/mol = 2341 g

5. Calculate the percent purity of Fe2O3 in the original sample:
- Percent purity = (Mass of Fe2O3 reacted / Mass of the original sample) * 100%
- Mass of the original sample = 2.62 kg = 2620 g
- Percent purity = (2341 g / 2620 g) * 100% = 89.30%

Therefore, the percent purity of Fe2O3 in the original sample is approximately 89.30%.