Question

10. Calculate the number of moles of hydrogen produced by the reaction of sodium with water, which resulted to formation of NaOH and H2? In the reaction, 1.3 L gas is collected by water displacement at 26 °C. The atmospheric pressure is 756 torr. (The partial pressure of water vapor at 26 °C is 25 torr.)

Answer 1

first calculate total mole of gas produced by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = atm pressure= 756 torr = 0.994737 atm,

V = volume in Liter = 1.3 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 26⁰C = 273.15+ 26 = 299.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.994737 1.3)/(0.08205 299.15) = 0.05268 mole of total gas produced

now calculate mole fraction of H₂

partial pressure of H₂ = 756 - 25 = 731 torr

X_i = P_i / P where, X_i =mole fraction of indivisual gas, P_i = partial pressure of indivisual gas, P = mole fraction

mole fraction of H₂ = 731/756 = 0.9669

no. of mole = mole fraction total mole

no. of mole of H₂ = 0.9669   0.05268 = 0.050936 mole of H₂ produced

0.050936 mole of H₂ produced