Question

How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s) + 3CO(g) -----à 3CO2(g) + 2Fe(s)

Answer 1

Answer : The grams of carbon monoxide needed are 148.89 g

Solution : Given,

Mass of iron, Fe = 198.5 g

Molar mass of iron, Fe = 56 g/mole

Molar mass of carbon monoxide, CO = 28 g/mole

First we have to calculate the moles of iron, Fe.

$\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles$

The balanced chemical reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)$

From the balanced reaction, we conclude that

2 moles of iron produces from the 3 moles of carbon monoxide

3.545 moles of iron produces from the $\frac{3}{2}\times 3.545=5.3175$ moles of carbon monoxide

Now we have to calculate the mass of carbon monoxide, CO.

$\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}$

$\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g$

Therefore, the grams of carbon monoxide needed are 148.89 g