n-butane is produced by monobromination of ethane followed by Wurtz's reaction. Calculate volume of ethane at NTP required to produce 55 g n-butane, if the bromination takes place with 90 % yield and the Wurtz's reaction with 85 % yield.

Question

n-butane is produced by monobromination of ethane followed by Wurtz's reaction. Calculate volume of ethane at NTP required to produce 55 g n-butane, if the bromination takes place with 90 % yield and the Wurtz's reaction with 85 % yield. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Reactions involved are C2H6 +Br2 arrow C2H5Br + HBr 2C2H5Br + 2Na arrow C4H10 + 2NaBr Actual yield of C4H10 = 55 g which is 85 % of theoretical yield. ⇒ Theoretical yield of C4H10 = ( 55 × 100/85 ) = 64 .70 g Also, 2 moles (218 g) C2H5Br gives 58 g of butane. ⇒ 64.70 g of butane would be obtained from (2/58) × 64.70 = 2.23 mole C2H5 Br Also yield of bromination reaction is only 90 %, in order to have 2.23 moles of C2H5Br , theoretically (2.23 × 100/90) = 2.48 moles of C2H5Br required. Therefore, moles of C2H6 required = 2.48 ⇒ Volume of C2H6 (NTP) required = 2.48 × 22.4 = 55.55 L.

Q. n-butane is produced by monobromination of ethane followed by Wurtz's reaction. Calculate volume of ethane at NTP required to produce 55gn-butane, if the bromination takes place with 90% yield and the Wurtz's reaction with 85% yield.

Q. $n$ -butane is produced by monobromination of ethane followed by Wurtz's reaction. Calculate volume of ethane at NTP required to produce $55 g n$ -butane, if the bromination takes place with $90%$ yield and the Wurtz's reaction with $85%$ yield.