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n-Butane is produced by the monobromination of ethane followed by word reaction .Calculate the volume of ethane at NTP to produce 55g n-butane if the bromination takes place with 90% yield and worth reaction with 85% yield.

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

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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.
answered Nov 6, 2013 by sreemathi.v
 

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