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

answer; natural resources;

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.