Question

004 1.0 points

What volume of oxygen gas can be collected
at 0.565 atm pressure and 26.0°C when 40.8
g of KCIO3 decompose by heating, according
to the following equation?
2KClO3———>2KCl +3O2
MnO2

Answer 1

V = 19.6 L

Step-by-step explanation:

Given data:

Mass of KClO₃ = 40.8 g

Pressure = 0.565 atm

Temperature = 26.0 °C ( 26+273= 299 K)

Volume of oxygen gas produced = ?

Solution:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

Number of moles of KClO₃:

Number of moles = mass/ molar mass

Number of moles = 40.8 g/ 122.55 g/mol

Number of moles = 0.3 mol

Now we will compare the moles of oxygen and KClO₃.

KClO₃ : O₂

2 : 3

0.3 : 3/2×0.2 = 0.45 mol

Volume of oxygen:

PV = nRT

V = nRT/P

V = 0.45 mol × 0.0821 atm. L. mol⁻¹. K⁻¹ . 299 K / 0.565 atm

V = 11.05 atm. L /0.565 atm

V = 19.6 L