Question

112 ml of a gas produced at STP by the action of 412 mg of alcohol ROH with excess CH3MgI. The Molecular mass of alcohol is
reaction:
CH ₃ Mg I +ROH _____ Ch4 + R - OMg-I

a) 32g b)41.2g c)82.4g d)156 g​

Answer 1

Answer:

option c = 82.4g

Explanation:

At STP one mole of a gas occupies 22.4 L

At STP 112ml is 1/200 moles hence looking at the reaction 1/200 moles of ROH gives 1/200  moles of CH4.

Given that 412 mg is required in this reaction means 1/200 moles of ROH is 412mg hence 1 mole = 412 ×200 mg

Hence the answer is C.

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