How much air is needed to completely burn 52 g of CH4 methane? (The air contains approximately 20% oxygen).
March 26, 2021
n (CH4) = 52g
M (O2) = 32
CH4 + 2O2 = CO2 + 2H2O
n (CH4) = m \ m = 52 \ 16 = 3.25 mol
m (O2) = 52 * 1 = 31 \ 16 = 208g
if there is 20% oxygen in the air, => air mass 208 \ 0.2 = 1040g
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