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 `(Al_(4)C_(3))` are required to produce 11.2 L of methane under STP conditions, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of aluminum carbide with water can be represented as follows:

\[ Al_4C_3 + 12H_2O \rightarrow 3CH_4 + 4Al(OH)_3 \]

### Step 2: Calculate 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, we can use the formula:

\[ \text{Number of moles of } CH_4 = \frac{\text{Volume of } CH_4}{22.4 \, \text{L/mol}} \]

Substituting the given volume:

\[ \text{Number of moles of } CH_4 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{moles} \]

### Step 3: Relate moles of methane to moles of aluminum carbide
From the balanced equation, we see that 1 mole of aluminum carbide produces 3 moles of methane. Therefore, to find the moles of aluminum carbide required to produce 0.5 moles of methane, we can set up the following proportion:

\[ \text{Moles of } Al_4C_3 = \frac{1 \, \text{mole } Al_4C_3}{3 \, \text{moles } CH_4} \times 0.5 \, \text{moles } CH_4 = \frac{0.5}{3} \, \text{moles } Al_4C_3 = \frac{1}{6} \, \text{moles } Al_4C_3 \]

### Step 4: Calculate the molar mass of aluminum carbide
To find the mass of aluminum carbide, we first need to calculate its molar mass:

\[ \text{Molar mass of } Al_4C_3 = (4 \times 27) + (3 \times 12) = 108 + 36 = 144 \, \text{g/mol} \]

### Step 5: Calculate the mass of aluminum carbide required
Now, we can use the number of moles calculated in Step 3 to find the mass:

\[ \text{Mass of } Al_4C_3 = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass of } Al_4C_3 = \left(\frac{1}{6} \, \text{moles}\right) \times 144 \, \text{g/mol} = 24 \, \text{g} \]

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

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