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Determine the volume of Oxygen gas produced by decomposition of 2.44g KClO3?

User Ian Smith
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1 Answer

16 votes
16 votes

Answer:

0.956 grams O₂ or
0.956 milliliters O₂ or
9.56 x 10⁻⁴ L O₂

Step-by-step explanation:

If the reactant were to decompose completely, the entire balanced reaction would look like this:

2 KClO₃ --> 2 K(s) + Cl₂(g) + 3 O₂(g)

To find the answer, (1) convert the grams of KClO₃ to moles (via molar mass from periodic table), then (2) convert moles KClO₃ to moles O₂ (via mole-to-mole ratio from balanced equation), then (3) convert moles O₂ to grams (via molar mass from periodic table).

Molar Mass (KClO₃) = 39.098 g/mol + 35.453 g/mol + 3(15.999) g/mol

Molar Mass (KClO₃) = 122.548 g/mol

Molar Mass (O₂) = 2(15.999) g/mol

Molar Mass (O₂) = 31.998 g/mol

2.44 g KClO₃ 1 mole KClO₃ 3 moles O₂ 31.998 g
-------------------- x -------------------------- x -------------------- x ------------------------
122.548 g 2 moles KClO₃ 1 mole O₂

= 0.956 grams O₂
= 0.956 milliliters O₂
= 9.56 x 10⁻⁴ L O₂

**The prompt askes to determine the volume. As far as I know, there is no easy way to find liters/milliliters with the given information. As such, I found the mass of O₂ instead. Technically, grams is equal to milliliters, so this may be what you are looking for. Then, you can convert from milliliters to liters by dividing by 1,000 (if necessary). **

User Ayan Biswas
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