How many moles of potassium chlorate were necessary to produce eight moles of oxygen?

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How many moles of potassium chlorate were necessary to produce eight moles of oxygen?

2 KClO > 2 KCl + 3O2

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Answer 1 · 2018-02-16T04:24:11

$5$ $5$ moles of potassium chlorate.

Explanation:

This is the equation that's been given:

$2 K C l O_{3} \to 2 K C l + 3 O_{2}$ $2KClO_3 -> 2KCl + 3O_2$

(The equation you wrote in the question was $2 K C l O \to 2 K C l + 3 O_{2}$ $2KClO -> 2KCl + 3O_2$ , but I think you meant $2 K C l O_{3} \to 2 K C l + 3 O_{2}$ $2KClO_3 -> 2KCl + 3O_2$ because you also wrote "potassium chlorate," which is $K C l O_{3}$ $KClO_3$ . Also because $2 K C l O_{3} \to 2 K C l + 3 O_{2}$ $2KClO_3 -> 2KCl + 3O_2$ is balanced.)

From looking at our balanced equation, we know that $2$ $2$ moles of potassium chlorate produces $3$ $3$ moles of oxygen, or $O_{2}$ $O_2$ .
So, our mole ratio of potassium chlorate to oxygen is $2 : 3$ $2:3$ .

To find out how much potassium chlorate is needed to produce $8$ $8$ moles of oxygen, we just need to take advantage of the fact that the value of the ratio will be the same when we multiply or divide both terms by the same number.

If we multiply $3$ $3$ by a number to get $8$ $8$ , the value of the ratio will be the same if we multiply $2$ $2$ by the same number! :)

$3 \times \frac{8}{3} = 8$ $3xx8/3 = 8$

$2 : 3 = 2 \times \frac{8}{3} : 3 \times \frac{8}{3} = 2 \times \frac{8}{3} : 8$ $2:3 = 2xx8/3:3xx8/3 = 2xx8/3:8$

Therefore, to produce $8$ $8$ moles of oxygen, $2 \times \frac{8}{3} = 5.33$ $2xx8/3=5.33$ moles of potassium chlorate is needed.
The question basically gave us only $1$ $1$ significant figure, so our answer would be $5$ $5$ moles of potassium chlorate.

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How many moles of potassium chlorate were necessary to produce eight moles of oxygen?

2 KClO > 2 KCl + 3O2

1 Answer

Explanation:

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