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A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20000 kJ of energy per day for cooking. How long will the cylinder last if the enthalpy of combustion, $\Delta H = - 2658kJ/mol$ for butane.
(A) 35 days
(B) 26 days
(C) 29 days
(D) 24 days

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Last updated date: 13th Jun 2024
Total views: 403.5k
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Answer
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Hint: The molecular formula of butane is ${C_4}{H_{10}}$. First, calculate the energy given by one mole of butane. Then calculate the total energy given by 11.2 kg of butane gas. From that energy, we can find the number of days the cylinder would last.

Complete step by step solution:
We will first calculate the energy given per mole of butane and then we will calculate the total energy we will obtain by the combustion of all the butane gas which is 11.2 kg. Then, we can find the number of the cylinder that will last.
- We know that the molecular formula of butane is ${C_4}{H_{10}}$. So,
Molecular mass of butane = 4(Atomic mass of C)+ 10(Atomic mass of H)
Molecular mass of butane = 4(12) + 10(1) = 58$gmmo{l^{ - 1}}$
- We are given that 1 mole of butane gives 2658 kJ energy.
So, we can write that 58gm of butane gives 2658 kJ energy, so 11200 gm of butane will give $\dfrac{{11200 \times 2658}}{{58}} = 513268.96kJ$
Here, we have written 58 gm of butane because that much weight is present in 1 mole of butane.
- So, it is given that 20000kJ is the energy needed for cooking per day. So, 513268.96kJ energy will be enough for $\dfrac{{513268.96 \times 1}}{{20000}} = 25.66 \approx 26$ days.
Thus, we can say that the cylinder will last for 26 days.

Therefore, the correct answer is (B).

Note: Remember that here the enthalpy change given is a negative value. This means that the energy is released by combustion of butane gas. The enthalpy change is positive in the reaction that absorbs the heat.