Question

(1.6): What volume of oxygen, at 18°C and 750 torr, can be
obtained from heating 100gm of potassium chlorate, KCIO,?
​

Answer 1

Answer:

Volume of oxygen obtained = 30 L

Explanation:

Given data:

Mass of KClO₃ = 100 g

Pressure  = 750 torr = 750/760 = 0.98 atm

Temperature  = 18°C = 18+273 = 291 K

Volume of oxygen obtained = ?

Solution:

Chemical equation:

2KClO₃    →     3O₂ + 2KCl

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 100 g/ 122.55 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of KClO₃ with oxygen.

              KClO₃            :               O₂

                  2                :                 3

                0.82            :              3/2×0.82 = 1.23 mol

Volume of oxygen:

PV = nRT

0.98 atm × V = 1.23 mol ×0.0821 atm.L/mol.K × 291 K

0.98 atm × V = 29.4 atm.L

V = 29.4 atm.L /0.98 atm

V = 30 L