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?
> 2kcl +3o2
mno2

V = 19.6 L

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

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