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3 years ago

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!

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