Limiting Reagent Aluminium and Oxygen JEE Chemistry Question

 

❓ Question

When 81 g of aluminium is allowed to react with 128 g of oxygen gas, find the mass of aluminium oxide $(Al_2O_3)$ produced.

(Molar masses: $Al=27\text{ g mol}^{-1}$, $O=16\text{ g mol}^{-1}$)

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✍️ Solution

The balanced chemical equation is:

$$4Al+3O_2\rightarrow2Al_2O_3$$

Moles of aluminium:

$$n(Al)=\frac{81}{27}=3\text{ mol}$$

Moles of oxygen:

$$n(O_2)=\frac{128}{32}=4\text{ mol}$$

From the balanced equation:

$$4\text{ mol Al}\rightarrow3\text{ mol }O_2$$

For $3$ mol Al, required $O_2$ is:

$$\frac{3}{4}\times3=\frac94=2.25\text{ mol}$$

Available oxygen is $4$ mol.

Hence, Al is the limiting reagent.

From the balanced equation:

$$4\text{ mol Al}\rightarrow2\text{ mol }Al_2O_3$$

Therefore,

$$3\text{ mol Al}\rightarrow\frac{2}{4}\times3$$
$$=1.5\text{ mol }Al_2O_3$$

Molar mass of $Al_2O_3$:

$$=2(27)+3(16)$$
$$=102\text{ g mol}^{-1}$$

Mass of $Al_2O_3$:

$$=1.5\times102$$
$$=153\text{ g}$$

✅ Final Answer

$$\boxed{153\text{ g}}$$