To solve this problem, we need to determine the product B formed from the given reaction sequence and then calculate the mass of B produced from 11.25 mg of chlorobenzene.
1. Identify the Reactions:
i) \( \text{Mg}, \text{dry ether} \): This reaction forms a Grignard reagent from chlorobenzene. \( \text{C}_6\text{H}_5\text{Cl} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{Mg}\text{Cl} \) (Phenylmagnesium chloride)
ii) \( \text{CO}_2, \text{H}_2\text{O}^+ \): This is the carboxylation of the Grignard reagent. \( \text{C}_6\text{H}_5\text{Mg}\text{Cl} + \text{CO}_2 \xrightarrow{\text{H}_2\text{O}^+} \text{C}_6\text{H}_5\text{COOH} + \text{Mg(OH)Cl} \) (Benzoic acid, product A)
iii) \( \text{NH}_3, \Delta \): This is the reaction of benzoic acid with ammonia under heat, forming benzamide (product B). \( \text{C}_6\text{H}_5\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2 + \text{H}_2\text{O} \) (Benzamide, product B)
2. Determine the Molecular Weights:
Chlorobenzene (\(\text{C}_6\text{H}_5\text{Cl}\)): (6 * 12) + (5 * 1) + 35.5 = 72 + 5 + 35.5 = 112.5 g/mol
Benzamide (\(\text{C}_6\text{H}_5\text{CONH}_2\)): (6 * 12) + (5 * 1) + 12 + 16 + 14 + 2 = 72 + 5 + 12 + 16 + 14 + 2 = 121 g/mol
3. Calculate Moles of Chlorobenzene:
Mass of chlorobenzene = 11.25 mg = 0.01125 g
Moles of chlorobenzene = 0.01125 g / 112.5 g/mol = 0.0001 mol
4. Determine Moles of Benzamide:
From the stoichiometry of the reactions, 1 mole of chlorobenzene produces 1 mole of benzamide. Therefore, moles of benzamide = 0.0001 mol.
5. Calculate Mass of Benzamide:
Mass of benzamide = moles of benzamide * molar mass of benzamide
Mass of benzamide = 0.0001 mol * 121 g/mol = 0.0121 g = 12.1 mg
6. Express the answer in the required format:
The mass of benzamide (B) is \(12.1 \text{ mg} = x \times 10^{-1} \text{ mg} \)
Therefore, \( 12.1 = x \times 10^{-1} \Rightarrow x = 121 \)
Final Answer:
The value of \( x \) is 121.
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