Question

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

Answer 1

`2.3molKClO_3`

`285.95gKClO_3`

Step-by-step 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*(2molKClO_3)/(3molO_2) \\\\2.3molKClO_3`

And in grams:

`2.3molKClO_3*(122.55gKClO_3)/(1molKClO_3) \\\\=285.95gKClO_3`

Regards!

Regards!