A. The barometric pressure is 766.1 mmHg. The vapor pressure of water at 24.0 C is 22.4 mmHg. Calculate the pressure, in mmHg, of the dry hydrogen gas.
B. Convert the pressure of the dry H2 to units of atm. 760.0 mm Hg = 1 atm.
C. Calculate moles of H2 produced in the reaction. R = 0.08206 L atm/(mol K)
According to dalton's law total pressure of gas is equal to sum of partial pressure exerted by each individual gas
Ptotal = P1 + P2
PH2 = Ptotal - Pwater vapour
PH2 = 766.1 - 22.4 = 743.7 mmHg
Pressure of dry hydrogen gas = 743.7 mmHg
760.0 mmHg = 1 atm then 743.7 mmHg = 743.7 X 1 / 760 = 0.97855 atm
743.7 mmHg = 0.97855 atm
Use ideal gas equation for calculation of mole of gas
Ideal gas equation
PV = nRT where, P = atm pressure= 0.97855 atm,
V = volume in Liter = 89.5 ml = 0.0895 L
n = number of mole = ?
R = 0.08206L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = 297.15 K
We can write ideal gas equation
n = PV/RT
Substitute the value
n = (0.97855X 0.0895) / (0.08206X 297.15) = 0.00359 mole
mole of H2 produced in reaction = 0.00359 mole
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