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

Volume of oxygen obtained = 30 L

Step-by-step 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