Question

Combining 0.400 mol of Fe2O3 with excess carbon produced 14.3 g of Fe.

Fe2O3 +3C ---> 2Fe + 3CO

1) What is the actual yield of Iron in moles?
2) What was the theoretical yield of iron in moles?
3) What was the percent yield?

Answer 1

Answer on Question #40482 - Chemistry - Other

Question

Combining 0.400 mol of $\mathrm{Fe_2O_3}$ with excess carbon produced 14.3 g of Fe.

$$\mathrm{Fe_2O_3} + 3\mathrm{C} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO}$$

1) What is the actual yield of Iron in moles?

2) What was the theoretical yield of iron in moles?

3) What was the percent yield?

Answer:

1) Number of moles equals:

$$n = \frac{m}{M}$$

m – Mass of the Iron, g.

M – Molar mass of Iron, M(Fe) = 55.85 g/mol.

Then number of moles of Iron produced by reaction is:

$$n(Fe) = \frac{m(Fe)}{M(Fe)} = \frac{14.3}{55.85} = 0.256 \text{ mol}$$

So, the actual yield of Iron in moles equals 0.256 mol.

2) Make a proportion:

$$1 \text{ mole of } \mathrm{Fe_2O_3} \text{ produces } 2 \text{ mol of Fe (Iron)}$$ $$0.400 \text{ moles of } \mathrm{Fe_2O_3} - x \text{ mol of Fe}$$ $$x = \frac{0.400 \cdot 2}{1} = 0.800 \text{ mol of Fe}$$

So, the theoretical yield of Iron in moles equals 0.800 mol.

3) The percent yield is the actual yield divided by the theoretical yield:

$$\% \text{yield} = \frac{0.256}{0.800} \times 100\% = 32.0\%$$

Answer: 1) 0.256 mol

2) 0.800 mol

3) 32.0 %