How many liters of oxygen gas at 153 °C and 0.820 atm can be prepared by the decomposition of 22.4 g of solid KCIO3?
2 KCIO3(s) > 2 KCl(s) + 3 O2(g)

V =  11.51 L

Explanation:

Given data:

Volume of oxygen gas produced = ?

Pressure = 0.820 atm

Temperature = 153°C (153+273 = 426K)

Mass of KClO₃ = 22.4 g

Solution:

Chemical equation:

2KClO₃    →         2KCl  + 3 O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles =  22.4 g/  122.5 g/mol

Number of moles = 0.18 mol

now we will compare the moles of KClO₃  and O₂

                       KClO₃           :              O₂

                           2               :               3

                         0.18             :             3/2×0.18 = 0.27

Volume of oxygen:

PV = nRT

R = general gas constant (0.0821 atm.L/mol.K)

0.820 atm× V = 0.27 mol × 0.0821 atm.L/mol.K ×426 K

V = 9.44 atm.L / 0.820 atm

V =  11.51 L

wher is the rest of the question