When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is_______ - (Nearest integer)
Given :
Molar mass of Al is $27.0 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of O is $16.0 \mathrm{~g} \mathrm{~mol}^{-1}$

Question

Aicharya · Accepted Answer

$\begin{aligned} & 4 \mathrm{Al}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{Al}_2 \mathrm{O}_3 \\ & \frac{81}{27}=3 \text { mole } \frac{128}{32}=4 \mathrm{~mole}\end{aligned}$
Limiting reagent
$\begin{aligned}
& \therefore \text { mole of } \mathrm{Al}_2 \mathrm{O}_3 \text { formed }=\frac{1}{2} \times 3 \text { mole } \\
& \therefore \text { wt. of } \mathrm{Al}_2 \mathrm{O}_3 \text { formed }=\frac{3}{2} \times 102 \\
& \quad=153 \mathrm{gm}
\end{aligned}$

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide…

Solution