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Alika [10]
3 years ago
9

How much reactant (KClO3) is required to produce 3.5 mil of O2

Chemistry
1 answer:
BigorU [14]3 years ago
3 0

Answer:

2.3molKClO_3

285.95gKClO_3

Explanation:

Hello there!

In this case, according to the balanced chemical equation:

2KClO_3\rightarrow 2KCl+3O_2

We can observe the 2:3 mole ratio in order to calculate the moles of KClO3 required for such production:

3.5molO_2*\frac{2molKClO_3}{3molO_2} \\\\2.3molKClO_3

And in grams:

2.3molKClO_3*\frac{122.55gKClO_3}{1molKClO_3} \\\\=285.95gKClO_3

Regards!

Regards!

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