a sample of iron are is known as Fe3O4. A 50 gram of sample contain 26 gram of Fe. Determine % yield Fe3O4 In sample

To determine the percent yield of Fe3O4 in the sample, we need to compare the actual yield (26 grams of Fe) with the theoretical yield of Fe3O4.

1. Calculate the molar mass of Fe and Fe3O4:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of Fe3O4 = (3*55.85) + (4*16.00) = 231.40 g/mol

2. Convert the mass of Fe to moles:
- Moles of Fe = mass of Fe / molar mass of Fe = 26 g / 55.85 g/mol = 0.465 mol

3. Use the stoichiometry of the balanced equation to determine the moles of Fe3O4:
- From the equation: 4 Fe + 3 O2 => 2 Fe2O3
- Since Fe3O4 represents a ratio of Fe to Fe2O3, we need to divide the moles of Fe by 4, then multiply by 2:
Moles of Fe2O3 = (0.465 mol Fe / 4) * 2 = 0.1163 mol Fe2O3

4. Calculate the mass of Fe3O4 from the moles:
- Mass of Fe3O4 = Moles of Fe2O3 * molar mass of Fe3O4
= 0.1163 mol * 231.40 g/mol
= 26.92 g

5. Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) * 100%
= (26 g / 26.92 g) * 100%
= 96.67%

Therefore, the percent yield of Fe3O4 in the sample is approximately 96.67%.