Answer to Question #261973 in Chemistry for Ban

Question

 Assuming that the oxygen gas is collected at P = 755 mm Hg and T = 305 K with the chemical equation 2KClO3(s) → 2KCl(s) + 3O2(g), how many liters of oxygen gas are produced when 294 g of KClO3 completely react as used in the ignition of fireworks? Solve to get the correct answer

Answer 1

Solution:

Calculate the moles of KClO₃:

The molar mass of KClO₃ is 122.55 g/mol

Hence,

(294 g KClO₃) × (1 mol KClO₃ / 122.55 g KClO₃) = 2.399 mol KClO₃ = 2.4 mol KClO₃

Balanced chemical equation:

2KClO₃ → 2KCl(s) + 3O₂(g)

According to stoichiometry:

2 mol of KClO₃ produces 3 mol of O₂

Thus, 2.4 mol of KClO₃ produces:

(2.4 mol KClO₃) × (3 mol O₂ / 2 mol KClO₃) = 3.6 mol O₂

The ideal gas law can be used: PV = nRT

To find the volume of O₂, solve the equation for V:

V = nRT/P

V(O₂) = (3.6 mol × 62.3637 L mmHg mol^-1 K^-1 × 305 K) / (755 mmHg) = 90.6958 L = 90.7 L

V(O₂) = 90.7 L

Answer: 90.7 liters of oxygen gas are produced.