A 100 mL solution of CH_3CH_2MgBr on treatme | vedclass

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(C) The reaction is: $CH_3CH_2MgBr + CH_3OH \rightarrow CH_3CH_3 + Mg(OCH_3)Br$.The gas produced is ethane $(C_2H_6)$.At $STP$,$22400 \ mL$ of gas cor

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$3$

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(C) The reaction is: $CH_3CH_2MgBr + CH_3OH ightarrow CH_3CH_3 + Mg(OCH_3)Br$.The gas produced is ethane $(C_2H_6)$.At $STP$,$22400 \ mL$ of gas corresponds to $1 \ mol$.Therefore,$2.24 \ mL$ of gas corresponds to $n = \frac{2.24}{22400} = 10^{-4} \ mol$.The molar mass of ethane $(C_2H_6)$ is $2 	imes 12 + 6 	imes 1 = 30 \ g/mol$.The weight of gas produced is $W = n 	imes M = 10^{-4} \ mol 	imes 30 \ g/mol = 30 	imes 10^{-4} \ g = 3 	imes 10^{-3} \ g$.Since $1 \ g = 1000 \ mg$,the weight is $3 	imes 10^{-3} 	imes 1000 = 3 \ mg$.

$A$ $100 \ mL$ solution of $CH_3CH_2MgBr$ on treatment with methanol produces $2.24 \ mL$ of a gas at $STP$. The weight of gas produced is $......mg$. [nearest integer]

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