Question

calculate mass of water vapour produce if 32gms of oxygen used in combustion.​

Answer 1

Answer:

look, there is a very simple way to calculate it.

Explanation:

CH4+2 O2= CO2+ 2H2O

32gms of methane is equal to 2 moles of methane as molecular we of methane is 16

Two moles of. Methane would produce 4 moles of water.A mole of water weighs 18 gms

Four moles of water would weigh 72 gm s

Therefore 32 gms of methane would produce 72 gms of water

Answer 2

Answer:

CH4+2 O2= CO2+ 2H2O

32gms of methane is equal to 2 moles of methane as molecular we of methane is 16

Two moles of. Methane would produce 4 moles of water.A mole of water weighs 18 gms

Four moles of water would weigh 72 gm s

Therefore 32 gms of methane would produce 72 gms of water

This is simplest answer assuming complete combustion of methane. If there is deficiency of oxygen combustion may not be complete and certain amount of carbon monoxide may form.Answer would be different and little complicated