Question

How many grams of iron oxide can be produced with 56.3 grams of o2 and 57.8 grams of fe


Basic explanations needed

Answer 1

The number of the moles of iron is:

$n=\frac{m}{M} = \frac{57.8\text{ g}}{ 55.845\text{ g/mol}} = 1.035 \text{ mol}$

The number of the moles of oxygen is:

$n=\frac{m}{M} = \frac{56.3\text{ g}}{ 32.00\text{ g/mol}} = 1.76 \text{ mol}$

Let's write the reaction equation and balance it:

4Fe + 3O₂ $\rightarrow$ 2Fe₂O₃

As you can see, 4 moles of iron react with 3 moles of oxygen. Therefore, the number of the moles of O₂ needed to react with all the iron available is: $1.035\cdot3/4 = 0.777$ mol. The amount of the oxygen is larger: 1.76 mol, so the oxygen is in excess. The number of the moles of the iron oxide Fe₂O₃ produced when all iron is consumed is:

$n = 1.035\cdot2/4 = 0.518$ mol

Therefore, its mass is:

$m = n\cdot M = 0.518 \text{ mol}\cdot159.69 \text{ g/mol} = 82.6$ g

Answer: 82.6 g of iron oxide can be produced with 56.3 grams of O₂ and 57.8 grams of Fe.