+2H2O 16gm2×22.4L22.4L 44gm No. of moles of CH4=
8
16
=0.5mol No. of moles of O2=
4.48L
22.4L
=0.2mol Now since 1 mole of CH4 requires 2 mol (i.e. 44.8L) of O2 for complete combustion. But the given moles of O2 is only 0.2mol.So,O2 is the limiting reagent. Again, since 2 moles of O2 reacts with 1mol of CH4 to give 22.4L of CO2 at STP. So 0.2 mole of O2 will react with 0.1mol of CH4 to give 2.24L of CO2. Wt. of CO2 produced =0.1mol×44=4.4gms of CO2