1.

A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20,000 kJ of energy per day for cooking, how long will the cylinder last if the enthalpy of combustion, Delta H = - 2658 kJfor butane?

Answer»

30.5 days
25.66 days
40.6 days
10.66 days

Solution :Given `C_4 H_10(g) + 13/2 O_2(g)to 4CO_2 (g) + 5H_2O (g) , Delta H = - 2658 kJ`
MOLECULAR weight of `C_4H_10 = 58 `g/mol
58 g of BUTANE on COMBUSTION PRODUCES = 2658 kJ heat
` therefore ` 11.2 kg of butane will produce
`= (2658)/(58) xx 11.2 xx 10^3 kJ` heat
= 513268.97kJ
The family needs 20,000 kJ of energy per day
` therefore ` Number of days ` = (513268.97)/(20000) kJ = 25.66 days`


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