12 g of Mg (at. Mass 24) on reacting completely with acid gives hydrogen gas, the volume of which at STP would be

Options

A. (a)22.4 L
B. (b)11.2 L
C. (c )44.8 L
D. (d)6.1 L

Explanation

To solve the problem of how much hydrogen gas is produced when 12 g of magnesium (Mg) reacts completely with acid, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the number of moles of magnesium (Mg)**:
- The atomic mass of magnesium (Mg) is given as 24 g/mol.
- To find the number of moles of magnesium in 12 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \, \text{g}}{24 \, \text{g/mol}} = 0.5 \, \text{moles} \]

2. **Identify the reaction**:
- The reaction of magnesium with an acid (like hydrochloric acid) can be represented as:
\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]
- From the balanced equation, we can see that 1 mole of Mg produces 1 mole of hydrogen gas (H₂).

3. **Calculate the moles of hydrogen gas produced**:
- Since 0.5 moles of magnesium react, the amount of hydrogen gas produced will also be 0.5 moles (from the stoichiometry of the reaction).

4. **Calculate the volume of hydrogen gas at STP**:
- At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters.
- Therefore, the volume of 0.5 moles of hydrogen gas is:
\[ \text{Volume} = \text{Number of moles} \times \text{Volume of 1 mole at STP} = 0.5 \, \text{moles} \times 22.4 \, \text{L/mole} = 11.2 \, \text{liters} \]

### Final Answer:
The volume of hydrogen gas produced at STP when 12 g of magnesium reacts completely with acid is **11.2 liters**.

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