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When 78.5gm of CH(3)COCl is treated with...

When `78.5gm` of `CH_(3)COCl` is treated with `29.75gm` of `CH_(3)MgBr` then what will be the mass of alcohol produces? (Given: `M`. wt. of `Cl=35.5, Br=80, C=12`)

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To solve the problem, we need to determine the mass of alcohol produced when 78.5 g of CH₃COCl is treated with 29.75 g of CH₃MgBr. We will follow these steps: ### Step 1: Calculate the molar mass of CH₃COCl - The molecular formula for acetyl chloride (CH₃COCl) can be broken down as follows: - C: 2 atoms × 12 g/mol = 24 g/mol - H: 3 atoms × 1 g/mol = 3 g/mol - O: 1 atom × 16 g/mol = 16 g/mol - Cl: 1 atom × 35.5 g/mol = 35.5 g/mol - Total molar mass of CH₃COCl = 24 + 3 + 16 + 35.5 = 78.5 g/mol ### Step 2: Calculate the moles of CH₃COCl - Given mass of CH₃COCl = 78.5 g - Moles of CH₃COCl = Given mass / Molar mass = 78.5 g / 78.5 g/mol = 1 mole ### Step 3: Calculate the molar mass of CH₃MgBr - The molecular formula for methyl magnesium bromide (CH₃MgBr) can be broken down as follows: - C: 1 atom × 12 g/mol = 12 g/mol - H: 3 atoms × 1 g/mol = 3 g/mol - Mg: 1 atom × 24 g/mol = 24 g/mol - Br: 1 atom × 80 g/mol = 80 g/mol - Total molar mass of CH₃MgBr = 12 + 3 + 24 + 80 = 119 g/mol ### Step 4: Calculate the moles of CH₃MgBr - Given mass of CH₃MgBr = 29.75 g - Moles of CH₃MgBr = Given mass / Molar mass = 29.75 g / 119 g/mol ≈ 0.25 moles ### Step 5: Determine the limiting reagent - The reaction between CH₃COCl and CH₃MgBr produces alcohol (CH₃CHOH) and the stoichiometry of the reaction is 1:1. - We have 1 mole of CH₃COCl and 0.25 moles of CH₃MgBr. - Since CH₃MgBr is the limiting reagent (less than 1 mole), it will determine the amount of alcohol produced. ### Step 6: Calculate the moles of alcohol produced - According to the stoichiometry of the reaction, 1 mole of CH₃MgBr produces 1 mole of alcohol. - Therefore, moles of alcohol produced = moles of CH₃MgBr = 0.25 moles. ### Step 7: Calculate the mass of alcohol produced - The molar mass of the alcohol (CH₃CHOH) is: - C: 2 atoms × 12 g/mol = 24 g/mol - H: 6 atoms × 1 g/mol = 6 g/mol - O: 1 atom × 16 g/mol = 16 g/mol - Total molar mass of alcohol = 24 + 6 + 16 = 46 g/mol - Mass of alcohol produced = moles × molar mass = 0.25 moles × 46 g/mol = 11.5 g ### Final Answer The mass of alcohol produced is **11.5 grams**. ---
To solve the problem, we need to determine the mass of alcohol produced when 78.5 g of CH₃COCl is treated with 29.75 g of CH₃MgBr. We will follow these steps: ### Step 1: Calculate the molar mass of CH₃COCl - The molecular formula for acetyl chloride (CH₃COCl) can be broken down as follows: - C: 2 atoms × 12 g/mol = 24 g/mol - H: 3 atoms × 1 g/mol = 3 g/mol - O: 1 atom × 16 g/mol = 16 g/mol - Cl: 1 atom × 35.5 g/mol = 35.5 g/mol ...
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