The minimum amount of air required for the complete combustion of 1L of water gas is 2.8L at NTP. Assuming that 20% excess air is used for the actual composition, calculate the volume of air used at 740 mm pressure and 25°C
n= 2.8 /22.4 = 0.125 moles of air. 0.125/0.8= 0.15625 moles of air with 20% excess.
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