Calculate the mass of KClO3 that will liberate 11.2 litre at NTP.
in grams
Your question is incomplete. It should be calculate the mass of KClO3 that will liberate 11.2 litres of O2 at NTP.
At STP , Volume of 1 mol = 24.8 L
2KClO3 → 2KCl + 3O2
Molar mass of KClO3 = 39 + 35.5 + 316 = 122.5 g
24.8 L of O2 produced at NTP = 122.5 g of KClO3
11.2 L of O2 produced = = 55.322 g of KClO3
At STP , Volume of 1 mol = 24.8 L
2KClO3 → 2KCl + 3O2
Molar mass of KClO3 = 39 + 35.5 + 316 = 122.5 g
24.8 L of O2 produced at NTP = 122.5 g of KClO3
11.2 L of O2 produced = = 55.322 g of KClO3