2
QuizMultiple Choice

Gas Stoichiometry — Quiz

Question 2 • Chemistry

Decomposition of potassium chlorate open paren KClO sub 3 close paren produces potassium chloride KCl and pure oxygen open paren cap o sub 2 close paren according to this balanced equation. 2 KCIO sub 3 s rightarrow 2 KCl s plus 3 cap o sub 2 g What volume of oxygen gas is released at STP if 10 point 0 g of potassium chlorate is decomposed? (The molar mass of KClO sub 3 is 122 point 5 5 g over mol )

Answer
A
1 point 8 3 liters
B
0 point 9 1 4 liters
C
2 point 7 4 liters
D
3 point 6 6 liters

Explanation

First convert potassium chlorate to moles: 10.0 g ÷ 122.55 g/mol = 0.0816 mol KClO₃. The balanced equation shows that 2 mol KClO₃ produce 3 mol O₂, so this makes 0.0816 × 3/2 = 0.122 mol O₂. At STP, 1 mol of gas occupies 22.4 L, giving 0.122 × 22.4 = 2.74 L. The answer is 2.74 liters of oxygen.

AI-written and checked against the verified answer.

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