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Potassium chlorate decomposes according the following BALANCED equation.

2 KClO3 → 2 KCl + 3O2

How many grams of oxygen gas would be produced if 33.0 g of potassium chlorate react?

User Kevin Bourrillion
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1 Answer

15 votes
15 votes

Answer:

12.9 g O₂

Step-by-step explanation:

To find the mass of oxygen gas produced, you need to (1) convert grams KClO₃ to moles KClO₃ (via molar mass from periodic table values), then (2) convert moles KClO₃ to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to match the given value (33.0 g).

Molar Mass (KClO₃): 39.098 g/mol + 35.45 g/mol + 3(15.998 g/mol)

Molar Mass (KClO): 122.542 g/mol

2 KClO₃ ---> 2 KCl + 3 O₂

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O): 31.996 g/mol

33.0 g KClO₃ 1 mole 3 moles O₂ 31.996 g
-------------------- x ------------------- x ----------------------- x ------------------ =
122.542 g 2 moles KClO₃ 1 mole

= 12.9 g O₂

User Ahmed Hamdy
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