To solve problems involving volumes of gases NOT at STP in chemical reactions: Combine PV = nRT and stoichiometry.

What volume of hydrogen will react w /carbon at 981 torr and 334 o C to yield 42.0 g of n-pentane? H 2 + CC 5 H g n-p () 1 mol n-p 72 g n-p P V = n R T = 135 = 3.5 mol H 2 L H 2 () 6 mol H 2 1 mol n-p

Dalton’s law of Partial Pressure P tot = P 1 + P partial pressures: total pressure of gaseous mixture pressure that each gas would exert by itself Other equations… Total moles of gas in a mixture: The mole fraction (X) of a gas in a mixture: n tot = n 1 + n 2 + …

Find the total pressure exerted by 38.0 g of carbon monoxide and 38.0 g of hydrogen in a 6.00-L container at 25 o C. n CO = 38 g () 1 mol 28.0 g 1. Find n tot. 2. Use PV = nRT to find P. = mol CO n H2 = 38 g () 1 mol 2.0 g = 19.0 mol H 2 n tot = mol = 8410kPa

With reference to the previous problem… What is the partial pressure exerted by each gas? P CO = () kPa= 560 kPa P H2 = () kPa= 7850 kPa The ratio of the partial pressures is the same as the mole ratio. mole fraction (38 g CO) (38 g H 2 )

Collecting Gases over Water filled w /water, initially before reaction gas being collected H 2 O levels even After rxn. is complete, raise or lower collecting vessel so H 2 O levels inside and out are the same. In this way... P atmos = P gas + P H2O during reaction reaction complete gas from reaction collected gas ( w /H 2 O vapor, too)

From bal. eq., n CaC2 = n C2H2, so g CaC2 = mol For the reaction CaC 2 (s) + H 2 O(l) C 2 H 2 (g) + CaO(s)... If L of acetylene are collected over water at 20.0 o C, find the moles of acetylene collected and the grams of calcium carbide used. The barometric pressure is torr torr = P C2H2 + P H2O o C (from p. 1111) So P C2H2 = torr = atm = mol C 2 H 2 () 1 mol 64.1 g = 2.13 g CaC 2

Find the total pres. in container Z, assuming constant T. PXPX VXVX VZVZ P X,Z A B C AB ZC 1.3 L 2.6 L 3.8 L 2.3 L 3.2 atm1.4 atm 2.7 atm X atm 3.2 atm1.3 L 2.3 L 1.81 atm 1.4 atm2.6 L1.58 atm 2.7 atm3.8 L4.46 atm 7.9 atm =