Question

If 0.905 mol Al2O3 is produced in the reaction, what mass of Fe in product's

Answer 1

Answer: Amount of iron produced is 101.36 grams.

Explanation:

For the reaction of aluminium and iron oxide, the equation follows:

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

By Stoichiometry of the reaction,

When 1 mole of aluminium oxide is produced, then 2 moles of iron is also produced.

So, when 0.905 moles of aluminium oxide is produced, then $\frac{ 2}{1}\times 0.905mol=1.81moles$ of iron is also produced.

Now, to calculate the amount of iron produced, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of iron = 56 g/mol

Moles of iron = 1.81 moles

Putting values in above equation, we get:

$1.81mol=\frac{\text{Mass of iron}}{56g/mol}\\\\\text{Mass of iron produced}=101.36g$

Hence, amount of iron produced is 101.36 grams.