Question

In the reaction, \[\ce{2 KClO_{3(S)} -> 2KCl_{(s)} + 3O_{2(g)} \Delta H°}\] = - 78kJ. If 33.6 L of oxygen gas is liberated at STP, what is the mass of KCl_(s) produced? (Atomic mass: K = 39, Cl = 35.5 g mol^-1)

Options

• 7.45 g

• 48.0 g

• 24.0 g

• 74.5 g

Answer 1

74.5 g

Explanation:

\[\ce{2 KClO_{3(S)} -> 2KCl_{(s)} + 3O_{2(g)}}\]

2 × 74.5 g       3 × 22.4 L

= 149 g          = 67.2 L

Now, 67.2 L of O₂ = 149 g of KCl at STP

33.6 L of O₂ = x g of KCl

∴ x = `(149 xx 33.6)/67.2` = 74.5 g of KCl