Butane C4H10, is a component of natural gas that is used for cigarette lighters.

Question

Butane C4H10, is a component of natural gas that is used for cigarette lighters. The balanced equation of the complete combustion of butane is:

2C4H10(g) + 13O2(g) ----> 8CO2(g)+10H2O(l)

At 1.00 atm and 23 degrees C what is the volume of carbon dioxide formed by the combustion of 2.20g of butane?

Explanation / Answer

molecular weight of butane = 58 gm 2C4H10(g) + 13O2(g) ----> 8CO2(g)+10H2O(l) 2 mole of butane produce 8moles of CO2 1 mole of butane produce 4moles of CO2 2.20 gm of butane = 0.03793 mole 0.03793mole of butane will produce CO2 =0.03793*4= 0.1517 mole co2 by gas law pv=nrt v= nrt/p n=0.1517 moles t=23+273=296 k p= 1atm = 101325 pa r = 8.314 units v=(0.1517* 8.314 *296)/101325 v=3.68* 10^-3 m3 v=3.68 L volume of CO2 ans

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