How much CO2 and how much water is formed during the combustion of 5g. dean?

dean – C10H22
2C10H22 + 31O2 = 20CO2 + 22H2O
amount of decane substance = 5g / (12 * 10 + 22) = 0.04 mol
1 mol takes 22.4 liters
n (co2) = 0.04 * 10 = 0.4
v (co2) = 0.4 * 22.4 = 0.896 liters
n (h20) = 0.04 * 11 = 0.44 mol
m (h20) = 0.44 * 18 = 7.92 g



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