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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 ? |
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Answer» Molecular mass of butane is = `58 g mol^(-1)` 58 g of butane on combustion gives 2658 kJ of heat `therefore` 11.2 g of butane on combustion gives `= (2658xx11.2xx1000)/(58)` kJ of heat The daily required of energy = 15000 kJ `therefore` The cylinder would last `=(2658xx11.2xx1000)/(58xx15000)` days `~~34` days |
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