$50$ % of the reagent is used for dehydrohalogentaion of $6.45 g m$ ${C H}_{3} {C H}_{2} C l$ . What will be the weight of the main product obtained?
(At. mass of $H, C$ and $C l$ are $1, 12$ and $35.5 g m / {m o l e}^{- 1}$ respectively)

0.7 g m

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1.4 g m

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2.8 g m

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5.6 g m

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Solution

The correct option is C $2.8 g m$
${C H}_{3} {C H}_{2} C l + K O H ⟶ C H_{2} = C H_{2} + K C l + H_{2} O$
Molar mass of ${C H}_{3} {C H}_{2} C l = 64.5 g .$
$64.5 g$ of ${C H}_{3} {C H}_{2} C l ⟹ \frac{1}{10}$ moles of $C 2 H_{5} C l .$
$⟹ \frac{1}{10}$ mol of $C H_{2} = C H_{2}$ produced.
Molar mass of $C_{2} H_{4} = 28 g ⟹ 2.8 g$ for $\frac{1}{10}$ mole.
Main product weight $= 2.8 g$

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