Answer to Question #62042 in General Chemistry for duna

Question

6. A 1.500 L flask is filled with a mixture of 1.20 g H2 and 8.40 g O2, at 25°C. The mixture is ignited, and hydrogen and oxygen combine to form water.
a. What is the total pressure inside the flask before the reaction? (Report your answer in units of atmospheres.)
b. What is the total pressure after the reaction, once the flask is returned to 25°C? (Vapour pressure of water at 25°C is 23.8 mm Hg.)

Answer 1

a. The total pressure of the mixture can be calculated as a sum of partial pressures of hydrogen and oxygen. Each pressure is divided as p = m*R*T/(M*V). Thus, p(H2) = 1.20 g * 0.082 L*atm/K/mol * 298 K / (2 g/mol * 1.500 L) = 9.8 atm. p(O2) = 8.40 g * 0.082 L*atm/K/mol * 298 K / (32 g/mol * 1.500 L) = 4.3 atm. The total pressure before ignition was 9.8 amt + 4.3 atm = 14.1 atm.
b. The burning follow the simple equation 2H2 + O2 = 2H2O. 2 mols of H2 needs 1 mol of O2 and 2 mols of water produced. We have 1.20 g/ 2 g/mol = 0.6 mols of H2 and 8.40 g/ 32 g/mol = 0.26 mol of O2. After burning it stays only 0.08 mol of H2 and 0.52 mol of H2O. 0.52 mol of H2O is some 9 g or 9 ml. Thus, the pressure of H2 after burning is 0.08 mol * 0.082 L*atm/K/mol * 298 K / (1.500 L - 0.009 L) = 1.31 atm. Together with pressure of water vapor the total pressure of gases after burning is 1.31 atm + 0.03 atm = 1.34 atm.