Question

A 367.8 g sample of potassium chlorate was decomposed according to the following equation:

2KClO3 -->2KCl + 3O2 How many moles of oxygen are formed?
a 2.5 mol
b 4.5 mol
c 1.0 mol
d 3.0 mol

Answer 1

When 367.8 g of potassium chlorate decomposes, 4.5 moles of oxygen gas are formed, based on stoichiometric calculations using the molar mass of potassium chlorate and the balanced chemical equation.

Step-by-step explanation:

To determine how many moles of oxygen gas (O₂) are formed when 367.8 g of potassium chlorate (KClO₃) decomposes, we need to use stoichiometry based on the balanced chemical equation:

2 KClO₃ (s) → 2 KCl (s) + 3 O₂ (g)

First, we find the molar mass of KClO₃:

1. K: 39.10 g/mol
2. Cl: 35.45 g/mol
3. O: 16.00 g/mol × 3 = 48.00 g/mol

Total molar mass of KClO₃ = 39.10 + 35.45 + 48.00 = 122.55 g/mol.

Next, we calculate the moles of KClO₃ in the 367.8 g sample:

367.8 g ÷ 122.55 g/mol = 3.0 moles of KClO₃.

Using the stoichiometric ratios from the balanced equation, for every 2 moles of KClO₃ decomposed, 3 moles of O₂ are produced:(3 moles KClO₃) × (3 moles O₂ / 2 moles KClO₃) = 4.5 moles O₂.

Thus, the answer is b. 4.5 mol of oxygen gas are formed.

Answer 2

the answer is A i did this before hope this helps