1. KEY
GENERAL CHEMISTRY-II (1412)
S.I. # 4
1. Define Dalton’s Law in words and equations.
In a gaseous mixture the total pressure is given by the sum of partial pressures of
each component: Ptotal = P1 + P2 + P3 + … where Pi = ni (RT/V) and
(Pi / PT) = (ni / nT) and Pi = Xi Ptotal where Xi is the mole fraction Xi = (ni /ntotal)
2. A mixture containing 2.50 g of each of CH4 (g), C2H4 (g) and C4H10 (g) is contained
in a 2.00 L flask at a temperature of 15°C. (a) Calculate the partial pressure of each
of the gases in the mixture (b) Calculate the total pressure of the mixture.
2.50 g CH4 (1mol / 16.04 g) = 0.15586 = 0.156 mol CH4
PCH4 = nRT / V  (0.156 mol)(0.0821 atmL)(288K) / 2 L = 1.84 atm
2.50 g C2H4 (1 mol / 28.05 g) = 0.0891 mol C2H4
PC2H4 = nRT/V  (0.0891 mol)(0.821 atmL)(288K) / 2L = 1.05 atm
2.50 g C4H10 (1 mol / 58.12 g) = 0.0430 mol C4H10
PC4H10 = nRT/V  (0.0430 mol)(0.821 atmL)(288K) / 2L = 0.508 atm
Ptotal = PCH4 + PC2H4 + PC4H10  1.84 atm + 1.05 atm + 0.508 atm = 3.40 atm
3. A mixture of gases contains 10.25 g of N2, 2.05 g of H2, and 7.63 g of NH3 g. If
the total pressure of the mixture is 2.35 atm, what is the partial pressure of each
component?
nN2 = 10.25 g (1 mol / 28.02 g) = 0.3658 mol
nH2 = 2.05 g (1 mol / 2.016g) = 1.0169 mol = 1.02 mol
nNH3 = 7.63 g( 1 mol / 17.03 g) = 0.448 mol
ntotal = 0.3658 + 1.02 + 0.448 = 1.8307 mol = 1.83 mol
PN2 = (nN2 / ntotal)(Ptotal) = (0.3658 / 1.8307)(2.35 atm) = 0.470 atm
PH2 = (nPH2 / ntotal)(Ptotal) = (1.0169/1.8307)(2.35 atm) = 1.31 atm
PNH3 = (nPNH3 / ntotatl)(Ptotal) = (0.448/1.8307)(2.35 atm) = 0.575 atm
4. Calcium hydride reacts with water to form hydrogen gas. Write a balanced
equation and determine how many grams of Calcium hydride are needed to
generate 53.5 L of H2 gas if the pressure of H2 is 814 torr at 21°C.

2. KEY
CaH2 (s) + 2 H2O (l)  Ca(OH)2 (aq) + 2 H2 (g)
nH2 = PH2V / RT  (814 torr/760torr) = 1.07 atm, T = 294K
nH2 = 1.07 atm (Kmol/ 0.0821 Latm)(53.5 L/ 294K) = 2.38 mol H2
(2.38 mol H2)(1mol CaH2 / 2 mol H2)(42.10 g CaH2 / 1 mol CaH2) = 50.0 g CaH2
5. What are the mole fractions of each component in a mixture of 5.08 g of O2, 7.17
g of N2, and 1.32 g of H2?
nO2 = (5.08 g)(1 mol / 32 g) = 0.159 mol
nN2 = (7.17 g)(1 mol / 28.02g) = 0.256 mol
nH2 = (1.32 g)(1 mol / 2.016g) = 0.655 mol
ntotal = 0.159 + 0.256 + 0.655 = 1.070 mol
Xo2 = nO2/nt = 0.159 / 1.07 = 0.149
XN2 = nN2/nt = 0.256 / 1.07 = 0.239
XH2 = nH2/nt = 0.655 / 1.07 = 0.612
6. Hydrogen gas is produced when zinc reacts with sulfuric acid:
Zn(s) H2SO4 (aq)  ZnSO4 (aq) + H2 (g)
If 159 mL of wet H2 is collected over water at 24°C and a barometric pressure of
738torr, how many grams of Zn have been consumed?
Pt = 738 torr = PH2 + PH2O Appendix B, V.P. of water at 24C is 22.38 torr
PH2 = (738 torr – 22.38 torr)(1/760 torr) = 0.942 atm
nH2 = PH2V / RT = (0.942 atm)(0.159 L) / (0.0821 Latm)(297 K) = 0.006143 =
0.00614 mol H2
(0.006143 mol H2)(1 mol Zn / 1 mol H2)(65.39 g Zn / 1 mol Zn) = 0.402 g Zn