What volume of O₂,measured at 225°C and 0.970 atm,will be produced by the decomposition of 3.16 g KClO₃?
(R = 0.08206 L · atm/mol · K) 2KClO₃(s) → 2KCl(s) + 3O₂(g)

Question

What volume of O₂,measured at 225°C and 0.970 atm,will be produced by the decomposition of 3.16 g KClO₃? (R = 0.08206 L · atm/mol · K) 2KClO₃(s)→ 2KCl(s)+ 3O₂(g)

Quizplus · Accepted Answer

First, we need to balance the chemical equation: 2KClO₃(s) → 2KCl(s) + 3O₂(g) Next, we need to use stoichiometry to find the moles of O2 gas produced: 3.16 g KClO₃ x (1 mol KClO₃/300.5 g KClO₃) x (3 mol O2/2 mol KClO₃) = 0.0158 mol O2 Finally, we can use the ideal gas law to find the volume of O2 gas produced: V = (nRT)/P = ((0.0158 mol)(0.08206 L·atm/mol·K)(225°C + 273.15 K))/(0.970 atm) = 1.63 L Therefore, the best choice is C.

What Volume of O2,measured at 225°C and 0

What Volume of O₂,measured at 225°C and 0