Answer:
V = 1.4 L
Explanation:
Given data:
Mass of KClO₃ = 5 g
Volume of oxygen produced = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl +3O₂
Number of moles of KClO₃:
Number of moles = mass/ molar mass
Number of moles = 5 g / 122.55 g/mol
Number of moles = 0.041
Now we will compare the moles of KClO₃ with oxygen,
KClO₃ : O₂
2 : 3
0.041 : 3/2× 0.041 = 0.062
Volume of oxygen at STP:
Standard pressure = 1 atm
Standard temperature = 273 K
PV = nRT
V = nRT/P
V = 0.062 × 0.0821 atm.L/mol.K × 273 K / 1 atm
V = 1.4 atm.L / 1atm
V = 1.4 L
