Hydrogen gas and oxygen gas are mixed in a 1.25 liter rigid reaction vessel so that the ratio of their pressures is three to one respectively. The temperature is 35.5 degrees Celsius and the total pressure is 1080 torr. A spark is introduced to combust the hydrogen. The final temperature is 80 degrees Celsius. The vapor pressure of water at 80 degrees Celsius is 355.1 torr. What reagent is the excess reagent and by how much?

Balanced chemical reaction is:-
2H2 + O2 ---------> 2H2O
Now, initial pressure of H2 = (3/4)*1080 = 810 torr
thus pressure of O2 = 1080 - 810 = 270 torr
Now, 760 torr = 1 atm
Thus, pressure of H2 = 1.066 atm ;
pressure of O2 = 270/760 = 0.355 atm
Now, applying Ideal Gas Equation, i.e P*V = n*R*T ; where P = pressure in atm ; V = volume in litres ; n = moles of the gas ; R = Universal gas constant = 0.0821 ; T = temperature in kelvin = 308.5 K (35.5 C)
Thus, moles of H2 present initailly = (1.066*1.25)/(0.0821*308.5) = 0.053
moles of O2 present initially = (0.355*1.25)/(0.0821*308.5) = 0.018

Now, as per the balanced reaction:-
H2 & O2 reacts in the molar ratio of 2:1
Thus, for 0.018 moles of O2 , moles of H2 required = 0.036
Clearly ,H2 is in excess and the excess amount is 0.053 - 0.036 = 0.017 moles
Now, molar mass of H2 = 2 g/mole
Thus, mass of excess H2 = 0.017*2 = 0.034 g