A cylinder of gas is assumed to contain 11.2 kg of butane ( C 4 H 10). If a normal family needs 20000 kJ of energy per day. The cylinder will last (Given that Δ H for combustion of butane is -2658 kJ )

Question

A cylinder of gas is assumed to contain 11.2 kg of butane ( C 4 H 10). If a normal family needs 20000 kJ of energy per day. The cylinder will last (Given that Δ H for combustion of butane is -2658 kJ ) (A) 20 days (B) 22 days (C) 26 days (D) 24 days

Tardigrade Admin · Accepted Answer

Cylinder contains 11.2 kg or 193.10 mole butane. ( ∵ mole mass of butane =58 ) ∵ Energy released by 1 mole of butane =-2658 ∴ Energy released by 193.10 mole of butane =-2658 × 193.10 =5.13 × 105 kJ ∴ (5.13 × 105/20000)=25.66 or 26 days

Q. A cylinder of gas is assumed to contain 11.2kg of butane (C4​H10​). If a normal family needs 20000kJ of energy per day. The cylinder will last (Given that ΔH for combustion of butane is −2658kJ )

20 days

22 days

26 days

24 days

Cylinder contains 11.2kg or 193.10 mole butane. ( ∵ mole mass of butane =58 ) ∵ Energy released by 1 mole of butane =−2658 ∴ Energy released by 193.10 mole of butane =−2658×193.10 =5.13×105kJ ∴200005.13×105​=25.66 or 26 days

Q. A cylinder of gas is assumed to contain $11.2 k g$ of butane $(C_{4} H_{10})$ . If a normal family needs $20000 k J$ of energy per day. The cylinder will last (Given that $Δ H$ for combustion of butane is $- 2658 k J$ )