2.4 g of pure Mg (at. Mass = 24) is dropped in 100 mL of 1 M HCl . Which of the following statement is wrong ?

Options

A. 1.12 L of hydrogen is produced at S.T.P
B. 0.05 mol of magnesium is left behind
C. HCl is the limiting reagent
D. 0.005 mol of magnesium is left behind

Answer

To solve the problem, we need to analyze the reaction between magnesium (Mg) and hydrochloric acid (HCl) and determine the correct statements based on the stoichiometry of the reaction.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between magnesium and hydrochloric acid can be represented as:
\[ \text{Mg} + 2 \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]

2. **Calculate the Moles of Magnesium (Mg):**
Given:
- Mass of Mg = 2.4 g
- Atomic mass of Mg = 24 g/mol

Using the formula for moles:
\[ \text{Moles of Mg} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.4 \, \text{g}}{24 \, \text{g/mol}} = 0.1 \, \text{mol} \]

3. **Calculate the Moles of Hydrochloric Acid (HCl):**
Given:
- Volume of HCl solution = 100 mL = 0.1 L
- Molarity of HCl = 1 M

Using the formula for moles:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \]

4. **Determine the Limiting Reagent:**
From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl. Therefore, for 0.1 moles of Mg, the required moles of HCl are:
\[ \text{Required HCl} = 2 \times 0.1 \, \text{mol} = 0.2 \, \text{mol} \]
Since we only have 0.1 moles of HCl available, HCl is the limiting reagent.

5. **Calculate the Moles of Hydrogen Gas (H2) Produced:**
From the balanced equation, 2 moles of HCl produce 1 mole of H2. Therefore, from 0.1 moles of HCl:
\[ \text{Moles of H2} = \frac{1}{2} \times 0.1 \, \text{mol} = 0.05 \, \text{mol} \]

6. **Calculate the Volume of Hydrogen Gas (H2) at STP:**
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Therefore, the volume of 0.05 moles of H2 is:
\[ \text{Volume of H2} = 0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

7. **Determine the Amount of Magnesium Left:**
Since 0.1 moles of Mg were present and 0.05 moles of Mg reacted with HCl, the remaining moles of Mg are:
\[ \text{Remaining Mg} = 0.1 \, \text{mol} - 0.05 \, \text{mol} = 0.05 \, \text{mol} \]

### Conclusion:
Now we can analyze the statements provided in the question:

1. **1.12 liters of hydrogen is produced at STP** - Correct.
2. **0.05 moles of Magnesium is left behind** - Correct.
3. **HCl is the limiting reagent** - Correct.
4. **0.005 moles of magnesium is left behind** - Incorrect (the correct amount is 0.05 moles).

### Final Answer:
The wrong statement is: **4. 0.005 moles of magnesium is left behind.**