At S.T.P. a mixture of 280 mL of CH_4 " and 140 mL of " H_2 iscompletely burnt. Calculate the required volume of oxygen and weight of water formed, assuming that whole of the steam condenses to water.

At S.T.P. a mixture of 280 mL of CH_4 " and 140 mL of " H_2 iscomplete

1.	At S.T.P. a mixture of 280 mL of CH_4 " and 140 mL of " H_2 iscompletely burnt. Calculate the required volume of oxygen and weight of water formed, assuming that whole of the steam condenses to water.
Answer» SOLUTION :For the combustion of methane : `{:(CH_4+2O_2to+2H_2O),(1 " mole " 2 " MOLES "2xx18=36g),(1"volume" 2" VOLUMES"):}` `:.1 "volume of " CH_4` requires for combustion, `O_2 = 2` volumes `:.280 " mL of "CH_4` will require for combustion, `O_2 = 2 xx 280 = 560 mL` Number of moles of `CH_4` in 280 mL at S.T.P. `=280/22400 =0.0125` `:. 1 " mole of " CH_4` gives on combustion, water = 36 G `:.0.0125 " mole of " CH_4` will give on combustion, water = `36/1 xx0.0125 = 0.45g` For the combustion of `H_2`: `{:(2H_2+O_2to2H_2O),(1 " mole " 2 " moles " 36g),(2 "volume " 1"volumes"):}` `:.2 " volumes of " H_2` require for combustion, oxygen = 1 volume `:. 140 mL H_2`will require for combustion, oxygen `=1/2 xx140 = 70 mL` Number of moles of `H_2` in 140 mL at S.T.P. `=140/22400 = 6.25 xx 10^(-3)` `:. 2 " moles of " H_2` produce water = 36 g `:. 6.25 xx 10^(-3) " moles of " H_2` will produce water `36/2xx6.25 xx 10^(-3)` `= 0.113 g` Hence, total volume of `O_2 " required " = 560 + 70 = 630 mL` and total mass of `H_2O " formed " = 0.45 +0.113 = 0.56 g`