$(a)\;56.3l\qquad(b)\;52.4l\qquad(c)\;55.53l\qquad(d)\;54l$

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Step 1:

$2C_2H_6+2Br_2\rightarrow 2C_2H_5Br+2HBr$

$2C_2H_5Br+2Na\rightarrow C_4H_{10}+2NaBr$

Weight of butane formed =55g

Mole of butane =$\large\frac{55}{58}$mole

Mole ratio in reaction :

$C_2H_6 : C_2H_5Br : C_4H_{10} \Rightarrow 2 : 2 :1$

$\therefore$ require Mole of $C_2H_5Br=\large\frac{55}{58}$$\times 2$

Since efficiency of worth reaction is 85%

Step 2:

Mole of $C_2H_5Br$ actually required $=\large\frac{55}{58}$$\times 2\times \large\frac{100}{85}$

Require mole of $C_2H_6=\large\frac{55}{58}$$\times 2\large\frac{100}{85}$

Given efficiency of $Br$ is 90%

$\therefore$ require mole of $C_2H_6$(actually)

$\Rightarrow \large\frac{55}{58}$$\times 2\times \large\frac{100}{85}\times \large\frac{90}{100}$

$\Rightarrow 2.479$ mole

$\therefore$ volume of $C_2H_6=22.4\times 2.479$

$\Rightarrow 55.53l$

Hence (c) is the correct answer.

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