JEE Chemistry 2023 Question with Solution

Given:

• Mass of magnesium $w = 2.4 \, \text{g}$
• Molar mass of magnesium $M = 24 \, \text{g/mol}$
• Molar volume of gas at STP = $22.4 \, \text{L}$

Find: The volume of hydrogen liberated at STP in the form _____ $\times 10^{-2}$ $\text{L}$

The balanced chemical equation is

$$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$$

Moles of magnesium are

$$n = \frac{w}{M} = \frac{2.4}{24} = 0.1$$

From the balanced equation, $1$ mole of $\text{Mg}$ produces $1$ mole of $\text{H}_2$. Therefore, moles of hydrogen gas produced are

$$0.1 \; \text{mol}$$

Now, volume of hydrogen at STP is

$$V = n \times 22.4 = 0.1 \times 22.4 = 2.24 \, \text{L}$$

So,

$$2.24 \, \text{L} = 224 \times 10^{-2} \, \text{L}$$

Therefore, the required numerical value is 224.

Given: Magnesium reacts with excess hydrochloric acid, so magnesium is the limiting reagent.

Find: The coefficient to fill in _____ $\times 10^{-2}$ $\text{L}$

Use the mole concept first. Magnesium has molar mass $24 \, \text{g/mol}$, so $24 \, \text{g}$ of magnesium corresponds to $1$ mole. Hence $2.4 \, \text{g}$ corresponds to

$$\frac{2.4}{24} = 0.1 \; \text{mol}$$

The reaction shows a $1:1$ mole ratio between $\text{Mg}$ and $\text{H}_2$, so hydrogen formed is also

$$0.1 \; \text{mol}$$

At STP, $1$ mole of any gas occupies $22.4 \, \text{L}$. Therefore,

$$0.1 \times 22.4 = 2.24 \, \text{L}$$

Expressing this as $\times 10^{-2} \, \text{L}$,

$$2.24 \, \text{L} = 224 \times 10^{-2} \, \text{L}$$

Thus, the answer is 224.