Answer to Question #109054 in General Chemistry for Jamie Strong

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.