What is the total pressure in a 6.00-l flask which contains 0.127 mol of h2 (g) and 0.288 mol of n2 (g) at 20.0°c?

Answer: 1.66 atm

dа ta:

P = ?

V = 6.00 liter

n1 = 0.127 mol of H2 (g)

n2 = 0.288 mol of N2 (g)

T = 20.0°C

Formula:

PV = n RT

Solution:

1) Conversion of units:

= 20 + 273.15 K = 293.15 K

2) n = n1 + n2 = 0.127 mol + 0.288 mol = 0.415 mol

3) PV = nRT = > P = nRT / V

R = 0.0821 atm*liter / (K*mol)

P = 0.415 mol * 0.0821 atm * liter / (K*mol) * 293.15 K / 6 liter = 1.66 atm

Answer: 1.66 atm