Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
