how much reactant (KClO3) is required to produce 3.5 mil of O2
![how much reactant KClO3 is required to produce 35 mil of O2 class=](https://us-static.z-dn.net/files/d4b/8ef45c5d8b2f8bb6da73522a4c8f957e.png)
Answer:
[tex]2.3molKClO_3[/tex]
[tex]285.95gKClO_3[/tex]
Explanation:
Hello there!
In this case, according to the balanced chemical equation:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
We can observe the 2:3 mole ratio in order to calculate the moles of KClO3 required for such production:
[tex]3.5molO_2*\frac{2molKClO_3}{3molO_2} \\\\2.3molKClO_3[/tex]
And in grams:
[tex]2.3molKClO_3*\frac{122.55gKClO_3}{1molKClO_3} \\\\=285.95gKClO_3[/tex]
Regards!
Regards!