Answer to Question #302413 in Chemistry for fairh

Question

if 146.5 g of chromium chlorite decomposes, what mass of oxygen can be produced

Answer 1

Solution:

chromium(III) chlorate = Cr(ClO₃)₃

Calculate the moles of Cr(ClO₃)₃:

The molar mass of Cr(ClO₃)₃ is 302.35 g/mol

Therefore,

Moles of Cr(ClO₃)₃ = (146.5 g) × (1 mol / 302.35) = 0.4845 mol

Balanced chemical equation:

2Cr(ClO₃)₃ → 2CrCl₃ + 9O₂

According to stoichiometry:

2 mol of Cr(ClO₃)₃ produces 9 mol of O₂

Thus, 0.4845 mol of Cr(ClO₃)₃ produces:

Moles of O₂ = [0.4845 mol Cr(ClO₃)₃] × [9 mol O₂ / 2 mol Cr(ClO₃)₃] = 2.18025 mol O₂

Calculate the mass of O₂:

The molar mass of O₂ is 32.0 g/mol

Therefore,

Mass of O₂ = (2.18025 mol O₂) × (32.0 g O₂ / 1 mol O₂) = 69.768 g O₂ = 69.77 g O₂

Mass of O₂ = 69.77 g

Answer: 69.77 grams of oxygen (O₂) can be produced.