Consider the reaction between 50.0 ml liquid methanol, ch3oh (density 0.850 g/ml), and 22.8 l o2 at 27c and a pressure of 2.00 atm. the products of the reaction are co2(g) and h2o(g). calculate the number of moles of h2o formed if the reaction goes to completion.

The reaction balanced equation is:
2CH3OH + 3O2 → 2CO2 + 4 H2O
the no.of moles of CH3OH = Mass of CH3OH / molar mass of CH3OH
when the mass of CH3OH = volume * density of CH3OH
                                             = 50 * 0.850 = 42.5 g
∴no of moles of CH3OH = 42.5 g / 32 g/mol =1.328 moles
after that to get the no.of moles of the gas O2 we have to decrease the T from 27 °C to zero and the pressure from 2 atm to 1 atm to get the corrected volume to STP.
So by substitution in this formula:
no.of moles of O2 = difference T in Kelvin * volume * change in pressure/molar volume
when we have difference T in kelvin to get V correction = 273K/(27°C+273K)=0.91 K
and Volume in the start = 22.8 L
the change in pressure = 2atm/1atm = 2 atm
and the molar volume = 22.4 L (as according to STP 1 mole of any gas occupies 22.4 L)
So by substitution:
no.ofmoles of O2 = (0.91 * 22.8 * 2) / 22.4 = 1.832 moles 
and according to the balanced equation:
∴ the no.of moles of H2O = no.ofmoles of O2 * (4 mol H2O /3 mol O2)
                                           = 1.832 * (4/3) =2.44 moles

A.