How many liters of oxygen gas are formed when 300 g of KClO3 completely react. Assume that the oxygen gas is collected at P =777mmHg and T =30°C. KClO3(s) ==> KCl(s) + O2(g) (put in sig fig)

Answer 1

KClO₃(s) ==> KCl(s) + O₂(g) is not balanced, so before we can proceed, we must balance the equation.

2KClO₃(s) ==> 2KCl(s) + 3O₂(g) ... balanced equation

First, we will find the mols of O2 produced from 300 g of KClO3:

molar mass KClO3 = 122.5 g/mol

300 g x 1 mol/122.5 g = 2.449 mols KClO3

Next, we will find mols of O2 produced from this:

2.449 mols KClO3 x 3 mols O2 / 2 mols KClO3 = 3.674 mols O2

Next, we will use the ideal gas law to find the volume of O2 produced from this many mols of O2:

P = 777 mm Hg x 1 atm / 760 mm Hg = 1.022 atm

30ºC + 273 = 303K

PV = nRT

V = nRT/P = (3.674 mols)(0.082 Latm/Kmol)(303K) / 1.022 atm

V = 89.43 L = 90 L (1 sig. fig. based on 1 sig. fig. in 300 g of KClO3)