[dylan]

Hello, I need help in finding the partial pressure of water vapour, part c, and am also unsure if my working out in the previous questions is correct.

QUESTION:
For a stoichiometric hydrogen-air reaction at 1atm pressure, find
(a) the fuel-to-mass ratio of f,
(b) the mass of fuel per mass of reactants, and
(c) the partial pressure of water vapour in the products

ATTEMPTED SOLUTION:
(a) eq. 2H2 + (O2 +3.76N2) --> 2H2O +3.76N2
 m(fuel)= 4* 1.008 g/mol = 4.032 g/mol
m(air)=2*16.00+7.52*14.01=137.35 g/mol
m(fuel)/m(air) =4.032/137.35 = 0.0294

(b) mf/mr => m(fuel)= 4.032 g/mol     and m(reactants)= 141.38 g/mol
m(fuel)/m(reactants) = 0.0285

(c) n=2 moles   m=36.032g
Am unsure as what to do next to find the partial pressure.

Thanks Dylan.