When heated, KCIO3 decomposes into KCl and O2
KCIO3 — 2 KCI + 3O2
If this reaction produced 51.4 g KCl, how many grams of O2were produced?

mass:

Explanation:

First, the KClO3 moles is determined by using given grams of KCl produced from the reaction.

Next, the moles of O2 produced is obtained by using the moles of KClO3 which then, finally converted into mass of O2.

Analyzing the given chemical reaction clearly shows that, 2 moles of KClO3 produces 2 moles of KCl and 3 moles of O2.

From the given grams of KCl produced the initial grams of KClO3 used in the reaction is determined as follows,

Explanation: this is a stoichiometry problem.

First, we convert grams of Kcl into moles by dividing by its molar mass:

(15.0 g KCl) x (1 moles/74.55g) = 0.201 moles KCl

Then, we convert moles of KCl to moles of O2 using the molar ratio from the balanced equation:

0.201 moles KCl x (3 moles O2/2 moles KCl) = 0.302 moles O2

Finally, we convert moles of O2 to grams of O2 by multiplying by its molar mass:

0.302 moles O2 x (32 g / 1mole) = 9.66 g O2

the answer is b. dilution