What volume of oxygen at STP can be produced by 6.125 g of potassium chlorate according to the reaction?
$2 K C l O_{3} \to 2 K C l + {3 O}_{2}$

4 L

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1.68 L

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3 L

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2.22 L

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Solution

The correct option is B 1.68 L
The number of moles of $K C l O_{3}$ is 6.125 g
So, $\frac{6.125}{122.5}$ = 0.05 moles.
Two molecules of $K C l O_{3}$ requires 3 molecules of $O_{2}$
Number of moles of $O_{2}$ = $3 \times 0.05 / 2$
= 0.075 moles
Volume of 1 mole = $22.4 L$
Volume of $O_{2}$ = 22.480.075
= 1.68 L
Therefore B is the correct option.

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