Question

1. A 50.0 L gas mixture at T = 500 K contains 123.0 g H2 and 855.0 g CO. What is the total pressure of the mixture? Assume ideal-gas behavior. Molar masses: M(H2) = 2.016 g/mol, M(CO) = 28.01 g/mol.

(A) 116 atm (B) 75.1 atm (C) 50.1 atm (D) 34.1 atm (E) 25.0 atm

2. KCl and K2SO4 are strong electrolytes. How do the concentrations of K+ ions, c(K+), in 2M KCl and 1M K2SO4 aqueous solutions compare?

(A) c(K+) is the same in both solutions.
(B) c(K+) in the KCl solution 2 times the concentration in the K2SO4 solution.

(C) c(K+) in the KCl solution 1/2 times the concentration in the K2SO4 solution.

(D) c(K+) in the KCl solution 4 times the concentration in the K2SO4 solution.

(E) c(K+) in the KCl solution 1/4 times the concentration in the K2SO4 solution.

3. Nb2O5 and Al can react as follows: 3Nb2O5(s) + 10Al(s) = 6Nb(s) + 5Al2O3(s). Theoretically, how many grams of Nb can a mixture of 11.60 g Nb2O5 and 2.60 g Al produce? Molar masses: M(Nb2O5) = 265.8 g/mol, M(Al) = 26.98 g/mol, and M(Nb) = 92.91 g/mol.

(A) 14.9 g (B) 8.95 g (C) 8.10 g (D) 5.37 g (E) 2.03 g

Answer 1

I moles of H₂ = 123g = 61.01 moles of co = 855g da 30.32 mol 28.01g/mol R = 0.0821 Lahm/malk H₂=nRT = = 50.09 atm 61.014 0.0821 x 500 50.OL Boko = MRT = 30.52 xO0821x500 = 25.05 atm 50.0 - Pt - PH2 + Pco = 50.09 +25.05=175.1 ahme kee kt+ce given imolofkt K2504 → 2K+ + 504? gives a mot of kt. c(kt) in K₂504 = 2M = 2X IM. clkt) in k ce=2M - Ixam. concentrations are same. A = 0.0436 mol 3. 3 N6₂ Os (5) + 10Al(s) 6 No(s) + 5 Al2O₃(s) moles of Nba Os = 11.69 265.8 g/mol moles of al = 2.6 g = 0.0963 mol (limiting) 26.98 g/mol : moles of Nb produced = 6 x 0.0963=0.057 mot mass ling) = 0.057* 92.91 = 5.3689 = 5.379 0