What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa) x (18.4 litres) / (8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 - -> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams