How many grams of oxygen, O2 are produced if 52.0g of potassium chlorate, KClO3, decomposes? (The molar mass of O2 = 32 g/mol and the molar mass of KClO3 = 122.4 g/mol).

20.39 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

Next, we shall determine the mass of KClO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of KClO3 = 122.4 g/mol

Mass of KClO3 from the balanced equation = 2 × 122.4 = 244.8 g

Molar mass of O2 = 32 g/mol

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

Summary:

From the balanced equation above,

244.8 g of KClO3 decomposed to produce 96 g of O2.

Finally, we shall determine the mass of O2 produced from the decomposition of 52 g of KClO3. This can be obtained as follow:

From the balanced equation above,

244.8 g of KClO3 decomposed to produce 96 g of O2.

Therefore, 52 g of KClO3 will decompose to produce =

(52 × 96)/244.8 = 20.39 g of O2.

Thus, 20.39 g of O2 is produced.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm