For the reaction represented by the equation 2KClO3 → 2KCl + 3O2, how many grams of potassium chlorate are required to produce 160. g of oxygen?

408.33g

Explanation:

The equation for the reaction is given below :

2KClO3 → 2KCl + 3O2

We'll begin by obtaining the mass of the KClO3 and O2 from the balanced equation. This can be achieved as shown below:

Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 39 + 35.5 + 48 = 122.5g/mol

Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

From the equation, 245g of KClO3 produced 96g of O2.

Therefore, Xg of KClO3 will produce 160g of O2 i.e

Xg of KClO3 = (245 x 160)/96 = 408.33g

Therefore, 408.33g of KClO3 are needed.