Question

2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3,Fe2O3, how many kilograms of Al2O3Al2O3 will be produced?

Answer 1

Answer : The mass of $Al_2O_3$ produced will be, 49.32 Kg

Explanation : Given,

Mass of $Al$ = 26.1 Kg  = 26100 g

Molar mass of $Al$ = 26.98 g/mole

Molar mass of $Al_2O_3$ = 32 g/mole

First we have to calculate the moles of $Al$.

$\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles$

Now we have to calculate the moles of $Al_2O_3$.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 moles of $Al$ react to give 1 mole of $Al_2O_3$

So, 967.38 moles of $Al$ react to give $\frac{967.38}{2}=483.69$ moles of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$.

$\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3$

$\text{Mass of }Al_2O_3=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg$

Therefore, the mass of $Al_2O_3$ produced will be, 49.32 Kg