A = CO₂; B = N₂

P_T = P_A + P_B

0.65 = P_A/P_T = P_A/90 atm = 58.5 atm = 59 atm

P_T = P_A + P_B

90 = 59 + P_B

P_B = 31 atm

b) NH₃ (g) + HF (g) ---> NH₄F (s)

a) MW NH₃ = 17.03 g/mol; MW HF = 20.01 g/mol

rate HF/rate NH₃ = (MW NH₃)^0.5/(MW HF) ^0.5

rate HF/rate NH₃ = (17.03 g/mol) ^0.5/(20.01 g/mol) ^0.5

rate HF/rate NH₃ = (4.1268)/(4.4729) = 0.9226

For every cm that NH₃ travels, HF travels 0.9226 cm

10.00 cm = x + 0.9226x

1.9226x = 10.00 cm

x = 10.00 cm/1.9226 = 5.201 cm

x = distance traveled by NH₃; the gases will meet 5.20 cm from the NH₃ flask.

MW = grams/moles, therefore find # of moles in the sample, and the corresponding # of grams

PV = nRT; (750 torr/760 torr/atm)(0.256 L) = n(0.08206 L*atm/mol*K)(373 K)

n = (0.9868 atm)(0.256 L)/(0.08206 L*atm/mol*K)(373 K) = 0.00825 mol gas in 256 mL

d = 3.125 g/L; (3.125 g/L)(.256 L) = 0.800 g

MW = 0.800g/0.00825 mol = 97.0 g/mol

EF = CHCl

weight = (12 + 1 + 35.5) g/mol = 48.5 g/mol

97.0/48.5 = 2

MF = C₂H₂Cl₂

2 NH₃ (g) + CO₂ (g) ---> H₂NC(O)NH₂ (s) + H₂O (g)

a) #mol NH₃:

n = PV/RT = (9.0 atm)(5.0 L)/(0.0821 L*atm/mol*K)(296 K) = 1.85 mol

#mol CO₂:

n = PV/RT = (14.0-9.0 atm)(5.0 L)/(0.0821 L*atm/mol*K)(296 K) = 1.03 mol

L.A.: (1.85 mol)/2 = 0.926 mol

(1.03 mol)/1 = 1.03 mol, therefore NH₃ is the limiting agent

#mol urea = (1.85 mol NH₃)(1 mol urea/2 mol NH₃)= 0.926 mol urea (H₂NC(O)NH₂)

MW H₂NC(O)NH₂ = 60.06 g/mol

mass urea = (0.926 mol)(60.06 g/mol) = 56 g urea

b) P = nRT/V

P = (0.926 mol)(0.0821 L*atm/mol*K)(373 K)/5.0 L = 5.7 atm

a) n = PV/RT

n = (40.0 atm)(2.0 L)/(0.0821 L*atm/mol*K)(273 K) = 3.6 mol

b) P = nRT/V

P = (3.6 mol)(0.0821 L*atm/mol*K)(318 K)/2.0 L = 47 atm

At the same temperature and pressure conditions, the volume is directly related to the number of moles.

1 Br₂ + 3 F₂ ---> 2 X

Mass Balance: 1 Br₂ + 3 F₂ ---> 2 BrF₃