Question

A 2.55 L sample of gas at STP contains hydrogen and oxygen gases. The mole fraction of hydrogen is 0.60. When the sample is combusted to form water, calculate the mass of water formed, assuming complete reaction.

Answer 1

at STP , total no of mole of gas mixer(n) = PV/RT

STP ,   T = 273 K , P = 1 atm

             R = 0.0821 l.atm.k-1.mol-1

         = 1*2.55/(0.0821*273)

           = 0.114 mole

so that,

No of mole of H2 = XH2*nTotal

                               = 0.6*0.114

                                = 0.0684 mole

No of mole of O2 in the mixer = 0.114 - 0.0684 = 0.0456 mole

2H2(g) + O2(g) ----> 2H2O(l)

2 mole H2 = 1 mole O2

so that,

No of mole of H2 = 0.0684 mole

No of mole of O2 in the mixer = 0.114 - 0.0684 = 0.0456 mole

limiting reactant in the mixer = H2

No of moleof water formed = 0.0684*2/2 = 0.0684 mole

mass of water = n*mwt = 0.0684*18 = 1.23g