Aluminum carbide `(Al_(4)C_(3))` liberates methane on treatment with water. The grams of aluminium carbide required to produce 11.2 L of methane under STP coniditions is [ Given Al =27]

Options

A. 48
B. 72
C. 144
D. 24

Answer

To solve the problem of how many grams of aluminum carbide `(Al₄C₃)` are required to produce 11.2 L of methane `(CH₄)` under STP conditions, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between aluminum carbide and water can be written as:
\[ \text{Al}_4\text{C}_3 + 12 \text{H}_2\text{O} \rightarrow 3 \text{CH}_4 + 4 \text{Al(OH)}_3 \]

### Step 2: Determine the moles of methane produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Therefore, to find the number of moles of methane produced from 11.2 L:
\[ \text{Moles of } CH_4 = \frac{\text{Volume}}{22.4 \text{ L/mol}} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \]

### Step 3: Relate moles of aluminum carbide to moles of methane
From the balanced equation, we see that 1 mole of aluminum carbide produces 3 moles of methane. Therefore, if we need 0.5 moles of methane, we can find the moles of aluminum carbide required:
\[ \text{Moles of } Al_4C_3 = \frac{0.5 \text{ moles } CH_4}{3} = \frac{1}{6} \text{ moles of } Al_4C_3 \]

### Step 4: Calculate the molar mass of aluminum carbide
The molar mass of aluminum carbide `(Al₄C₃)` can be calculated as follows:
- Aluminum (Al) has a molar mass of 27 g/mol.
- Carbon (C) has a molar mass of 12 g/mol.

Thus, the molar mass of aluminum carbide is:
\[ \text{Molar mass of } Al_4C_3 = (4 \times 27) + (3 \times 12) = 108 + 36 = 144 \text{ g/mol} \]

### Step 5: Calculate the mass of aluminum carbide required
Using the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
Substituting the values we have:
\[ \text{Mass of } Al_4C_3 = \frac{1}{6} \text{ moles} \times 144 \text{ g/mol} = 24 \text{ grams} \]

### Final Answer
The grams of aluminum carbide required to produce 11.2 L of methane under STP conditions is **24 grams**.

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