13.3g of an alkyl magnesium bromide reacts with an alcohol to give on alkane, which occupies a volume of 2.24 litre at STP. Find the number of carbon atoms present in the alkane.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the moles of Alkyl Magnesium Bromide (R-MgBr)
We are given the mass of the alkyl magnesium bromide (R-MgBr) as 13.3 g. To find the number of moles, we need the molar mass of R-MgBr.

**Molar mass of R-MgBr:**
- Molar mass of Magnesium (Mg) = 24 g/mol
- Molar mass of Bromine (Br) = 80 g/mol
- Let the molar mass of the alkyl group (R) = Mr

Thus, the molar mass of R-MgBr = Mr + 24 + 80 = Mr + 104 g/mol.

**Moles of R-MgBr:**
Using the formula for moles:
\[ \text{Moles of R-MgBr} = \frac{\text{mass}}{\text{molar mass}} = \frac{13.3 \text{ g}}{Mr + 104 \text{ g/mol}} \]

### Step 2: Calculate the moles of alkane produced
We know that the volume of the alkane produced is 2.24 liters at STP. At STP, 1 mole of any gas occupies 22.4 liters.

**Moles of alkane:**
\[ \text{Moles of alkane} = \frac{\text{Volume}}{22.4 \text{ L/mol}} = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ moles} \]

### Step 3: Relate moles of R-MgBr to moles of alkane
From the reaction of R-MgBr with alcohol, we know that:
- 1 mole of R-MgBr produces 1 mole of alkane (R-H).

Thus, the moles of R-MgBr will be equal to the moles of alkane:
\[ \frac{13.3 \text{ g}}{Mr + 104} = 0.1 \]

### Step 4: Solve for Mr (Molar mass of R)
Rearranging the equation:
\[ 13.3 = 0.1 \times (Mr + 104) \]
\[ 13.3 = 0.1Mr + 10.4 \]
\[ 0.1Mr = 13.3 - 10.4 \]
\[ 0.1Mr = 2.9 \]
\[ Mr = \frac{2.9}{0.1} = 29 \text{ g/mol} \]

### Step 5: Determine the number of carbon atoms in R
Assuming R is an alkyl group (CnH2n+1), we can find the number of carbon atoms (n) using the molar mass of carbon (C = 12 g/mol).

Let’s assume R = CnH2n+1:
\[ Mr = 12n + 1 \text{ (for H)} = 29 \]
Thus,
\[ 12n + 1 = 29 \]
\[ 12n = 28 \]
\[ n = \frac{28}{12} \approx 2.33 \]

Since n must be a whole number, we round it down to 2, indicating that there are 2 carbon atoms in the alkyl group.

### Conclusion
The number of carbon atoms present in the alkane is **2**.

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