If 59.3 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced?

1 answer:

8 0

Moles of KClO₃ = 59.3 / (39 + 35.5 + 16 x 3)
Moles of KClO₃ = 0.48

Molar ratio of KClO₃ to O₂ = 2 : 3

Moles of O₂ produced = 1.5 x 0.48
= 0.72

Mass of O₂ = moles x Mr
= 0.72 x 32
= 23.04 grams

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Answer:

<h3>The answer is 4.65 moles</h3>

Explanation:

To find the number of moles given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 2.8 × 10²⁴ atoms of Cl2

So we have

$n = \frac{2.8 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 4.65116279069...$

We have the final answer as

<h3>4.65 moles</h3>

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