Problems
1) If a gas tube in a closed tube manometer has a pressure of 0.067 atm, what is its pressure in mm Hg?
2) If you have a sample of 2.41 L of neon gas at 21 degrees C and a pressure of 0.963 atm and you increase the pressure to 1.26 atm but hold the temperature constant, what volume would the sample of neon gas occupy?
3) What volume would 2.85 mol chlorine gas occupy at 21 degrees C and 0.950 atm? (assume ideal behavior)
4) Given the following chemical reaction:
CaC2(s) + 2H2O(l) -> Ca(OH)2 (aq) + C2H2(g)
How many liters of acetylene gas, C2H2 are produced at 25 degrees C and 715 mm Hg from 0.0725 mol CaC2 in an excess of H2O?
5) A gas mixture contains 4.5%C, 70%N and 25.5%O. What are the partial pressures of each of these gases if the total pressure is 1 atm?
Answers
1) 0.067 atm x (760 mm Hg/1 atm) = 50.92 mm Hg
2) P1V1 = P2V2, V2 = P1V1/P2 = (0.963atm)(2.41L)/(1.26atm) = 1.81 L
3) PV = nRT, V = nRT/P = (2.85mol)(0.0821 L*atm/K*mol)(21+273K)/(0.950atm) = 72.4 L
4) 0.0725 mol CaC2 x (1 mol C2H2/1 mol C2H2) = 0.0725 mol C2H2
715 mm Hg x (1atm/760 mmHg) = 0.941 atm
PV = nRT, V = nRT/P = (0.0725 mol)(0.0821 L*atm/K*mol)(25+273K)/(0.941atm) = 1.9L
5) Assume you have a 100 gram sample, thus you have 4.5 grams C, 70 grams N and 25.5 grams O. Calculate the number of moles and determine the mole fraction for each element since partial pressure is defined by the equation Pi = XiPt.
4.5 grams C x (1 mol C/12.0 g) = 0.38 mol C
70 grams N x (1mol N/14.0g) = 5 mol N
25.5 grams O x (1 mol O/16.0g) = 1.6 mol O
Thus the mole fraction for each will be equal to the moles of each element divided by the total moles. Then multiplied by the total pressure to calculate the partial pressure each gas exerts.
Total number of moles = 6.98 moles
Partial pressure C = (.38mol/6.98mol)(1 atm) = 0.054 atm
Partial pressure N = (5 mol/6.98mol)(1atm) = 0.72 atm
Partial pressure O = (1.6 mol/6.93mol)(1atm) = 0.23 atm