Chemistry - Aromatic Hydrocarbon Question with Solution | TestHub

The question asks about the volume of benzene produced from the reaction of phenyl magnesium iodide with ethyne. Let's break down the solution step-by-step.

1. Write the balanced chemical equation:

$$2\mathrm{PhMgI}+\mathrm{CH}\equiv \mathrm{CH}⟶2\mathrm{PhH}+(\mathrm{C}\equiv \mathrm{C}{)}^{2-}({\mathrm{MgI}}^{+}{)}_2$$

Here, Ph represents the phenyl group (${\mathrm{C}}_6{\mathrm{H}}_5$). Phenyl magnesium iodide (PhMgI) reacts with ethyne (acetylene, ${\mathrm{C}}_2{\mathrm{H}}_2$) to produce benzene (PhH or ${\mathrm{C}}_6{\mathrm{H}}_6$) and a magnesium iodide acetylide.

2. Calculate moles of phenyl magnesium iodide (PhMgI):

The molecular formula of phenyl magnesium iodide is ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{MgI}$. Its molar mass is:

$6\times 12.01+5\times 1.01+24.31+126.90=72.06+5.05+24.31+126.90=228.32g/mol$

Given mass of PhMgI = 2.28 g

Moles of PhMgI = $\frac{2.28\mathrm{g}}{228.32g/mol}\approx 0.01\mathrm{mol}$

3. Calculate moles of ethyne (${\mathrm{C}}_2{\mathrm{H}}_2$):

Given volume of ethyne = 112 cc = 112 mL = 0.112 L

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Moles of ${\mathrm{C}}_2{\mathrm{H}}_2$ = $\frac{0.112\mathrm{L}}{22.4L/mol}=0.005\mathrm{mol}$

4. Determine the limiting reactant:

From the balanced equation, 2 moles of PhMgI react with 1 mole of ${\mathrm{C}}_2{\mathrm{H}}_2$.

We have 0.01 mol of PhMgI and 0.005 mol of ${\mathrm{C}}_2{\mathrm{H}}_2$.

The mole ratio required is $\frac{\mathrm{PhMgI}}{{\mathrm{C}}_2{\mathrm{H}}_2}=\frac{2}{1}$.

The actual mole ratio is $\frac{0.01}{0.005}=\frac{2}{1}$.

Since the mole ratio matches the stoichiometry, neither reactant is in excess.

5. Calculate moles of benzene produced:

From the balanced equation, 2 moles of PhMgI produce 2 moles of benzene (PhH). Also, 1 mole of ethyne produces 2 moles of benzene. Therefore, moles of benzene produced = moles of PhMgI = 0.01 mol. Alternatively, moles of benzene produced = 2 * moles of ethyne = $2\times 0.005=0.01$ mol.

6. Calculate volume of benzene produced at STP:

Volume of benzene = moles $\times$ molar volume at STP

Volume of benzene = $0.01\mathrm{mol}\times 22.4L/mol=0.224\mathrm{L}$

Therefore, the volume of benzene produced is 0.224 L.

FINAL ANSWER: D