A) what is the molar fraction of each gas in a mixture of 6g of hydrogen H2 and 7g of nitrogen N2 ?

Question

B) what is the total pressurd exerted by the above mixture in a volume of 15 l at a temperature of 0ºc ?
C) what is the partial pressure of each gas in the above mixture?
R=8.314J.mol.k

Answer 1

a.
mols H2 = g/molar mass = ?
mols N2 = g/molar mass = ?
XH2 = mols H2/total mols
XN2 = mols N2/total mols

b.
PV = nRT where n is total mols. P will be in kPa if R is 8.314 and that should be J/mol.K

c.
pH2 = XH2*Ptotal
pN2 = XN2*Ptotal

Answer 2

A. Mole Fraction (X₁) = [ moles of substance(n₁) / Total moles of all substances (∑n)
Moles H₂ = (6 gms/ 2 gms/mol) = 3 moles
Moles N₂ = (7 gms/ 14 gms/mol) = 0.50 mole
∑moles = (3 + 0.50)moles = 3.5 moles
X(H₂) = [(moles H₂)/(moles H₂ + moles N₂)] = 3/(3.0 + 0.50) = 3/3.5 = 0.8571
X(N₂) = 1 – X(H₂) = 1 – 0.8571 = 0.1486

B.Given 15.0 Liters total Volume at 0⁰C (=273K); use the Ideal Gas Law to calculate the partial pressures of H₂ and N₂.
PV = nRT => P = nRT/V
For H₂: n = 3 moles
For N₂: n = 0.50 mole
R = Constant = 0.08206 L-Atm/mol-K
T = 0⁰C = 273K
V(Total) = 15 Liters

(The following answer C-part)
P(H₂) = (3 moles)(0.08206 l-atm/mol-K)(273K)/(15 L) = 4.48 Atm x 760 mm-Hg/Atm = 3405 mm-Hg

P(N₂) = (0.50 mole)(0.08206 l-atm/mol-K)(273K)/(15 L) = 0.747 Atm x 760 mm-Hg/Atm = 568 mm-Hg

(The following answer B-Part)
Total Pressure = P(H₂) + P(N₂) = 4.48 Atm + 0.747 Atm = 5.23 Atm x 760 mm-Hg/Atm =3973 mm-Hg