A cylinder of cooking gas is assumed to contain 11.2 kg of butane. The thermo chemical equation for the combustion of butane is : `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)rarr 4CO_(2)(g)+5H_(2)O(l) , Delta H = -2658 kJ`
If a family needs 15000 kJ of energy per day for cooking, how would the cylinder last ?

A cooking gas cylinder is assumed to contain 11.2kg isobutane. The combustion of isobutane is given by : C_(4)H_(10(g))+(13//2)O_(2(g)) rarr 4CO_(2(g))+5H_(2)O_((l)), DeltaH=-2658kJ mol^(-1) (a) If a family needs 15000kJ of energy per day for cooking, how long would the cylinder last ? (b) Assuming that 30% of the gas is wasted due to incomplete combustion, how long would the cylinder last ?

A cylinder of cooking gas is assumed to contain 14.2 kg butane (C_(4)H_(10)) combustion of butane is C _(4) H _(10 (g)) + (13)/(2) O _(2 (g)) to 4CO _(2 (g)) + 5 H _(2) O _((l)) , Delta H =- 1660 kJ If a family needs 14000 kJ of energy per day for cooking and assuming that 20% of the gas is wasted due to incomplete combustion, how long would the cylinder last?

C_(4)H_(10)+(13)/(2)O_(2)(g)rarr4CO_(2)(g)+5H_(2)O(l), DeltaH =- 2878 kJ DeltaH is the heat of……….of butane gas.

C_(4)H_(10)+(13)/(2)O_(2)(g)rarr4CO_(2)(g)+5H_(2)O(l), DeltaH =- 2878 kJ DeltaH is the heat of……….of butane gas.

For the reaction C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2) (g) +2H_(2)O(l) , Delta E=-1415 kJ . The DeltaH at 27^(@)C is

A fuel cell involves combustion of butane at 1 atm and 298 K C_(4)H_(10 )(g)13/2 O_(2) (8)rarr 4CO_(2)(g)+5H_(2)0 (l), triangle G^(@) = -2746 kJ/mol. What is E^(@) of a cell?

For the reaction : C_(2)H_(4)(g) + 3 O_(2)(g) rarr 2 CO_(2)(g) + 2 H_(2)O(l) at 298 K, Delta U = - 1415 kJ . If R = 0.0084 kJ K^(-1) . Then Delta H is equal to