[living2434]

A mixture of hydrogen gas, oxygen gas, and 2 mL of water (l) is present in a 0.500L rigid container at 25 degrees Celsius. The number of moles of hydrogen gas and the number of moles of oxygen gas are equal. The total pressure is 1146 mm Hg. (The equilibrium vapor pressure of pure water at 25 degrees Celsius is 24 mm Hg). The mixture is sparked, and hydrogen gas and oxygen gas react until one reactant is completely consumed.

a. Identify the reactant remaining and calculate the amount (in moles) of this reactant remaining.
b. Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius. (The equilibrium vapor pressure of water at 90 degrees Celsius is 526 mm Hg)
c. Calculate the amount (in moles) of water present as vapor in the container at 90 degrees Celsius.

(My ANSWERS)
a.

n total  = PV/RT
           = (152.8kPa)(0.500L)/(8.314)(298K)
           = 0.0308mol

n water = PV/RT
           = (3.2kPa)(0.500L)/(8.314)(298K)
            = 0.00065mol

n gases = n total - n water
           = 0.0308mol - 0.0065mol
           = 0.0302mol

The ratio between reactants is 2H₂ : 1O₂
since # of mols for both reactants r the same, divide 0.302mol by 2 to get n for each gas
Coefficients tell us that H₂ is the LR
n then i divided 0.302mol by 4 to get the amount left in oxygen gas after H₂ totally reacts which then i get
0.00755mol as my final answer

b.
p = nRT/V
   = (0.0308mol)(8.314)(363K) / (0.500L)
    = 186 kPa

P total = p + p water at 90 degrees
          = 186 kPa + 70.1 kPa
          = 256kPa

c.
n = PV/RT
    = (70.1kPa)(0.002L)/(8.314)(363K)
    = 0.00005 mol

My answers are rounded according to sig dig rules

Thx for helping

[Borek]

a looks fine.