Question

Use the References to access important values if needed for this question. Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KCIO3(s)— 2KCI(s) + 302(g) The product gas, 02, is collected over water at a temperature of 25 °C and a pressure of 747 mm Hg. If the wet O2 gas formed occupies a volume of 6.24 L, the number of grams of O2 formed is g. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

Pressure of O2 = 747 - 23.8 => 723.2 mm of Hg

= 723.2/760 atm

Temperature= 298 K

Volume = 6.24 L

Using ideal gas equation

PV = nRT

n = 6.24×723.2/760×0.0821×298

=> 0.243 mol

Mass =0.243×32 => 7.77 g

2) pressure of O2 = 751-23.8 => 727.2 mm of Hg

PV = nRT

n = 9.09×727.2/760×0.0821×298

= 0.359 mol

Moles of KClO3 reacts = 2×0.359/3

=> 0.239 mol