Answer to Question #111343 in General Chemistry for Alana M

Question

The following reaction forms 6.41 L of oxygen at a temperature of 18.7oC and a pressure of 731 torr, what mass (in grams) of KClO3 must have decomposed? Round your answer to three sig figs.

2KClO3 (s) --> 2KCl (s) + 3O2 (g)

Answer 1

P = 731 torr = 731 torr * (1 atm / 760 torr) = 0.9618 atm

T = 18.7^oC = (18.7^oC + 273.15) = 291.85 K

R = 0.08206 L atm K^-1 mol^-1

Solution:

Balanced equation for the given chemical reaction:

2KClO₃(s) --> 2KCl(_s)+ 3O₂(_g)

Let's use the ideal gas equation to find the moles of O₂ that form.

The ideal gas equation can be expressed as: PV = nRT.

n = PV / RT

Then,

n(O₂) = (0.9618 atm * 6.41 L) / (0.08206 L atm K^-1 mol^-1 * 291.85 K) = 0.2574 moles

Moles of O₂ = n(O₂) = 0.2574 moles

According to the chemical equation: n(KClO₃)/2 = n(O₂)/3.

n(KClO₃) = 2*n(O₂)/3;

n(KClO₃) = (2 * 0.2574 moles) / 3 = 0.1716 moles.

Moles of KClO₃ = n(KClO₃) = 0.1716 moles.

Moles of KClO₃ = Mass of KClO₃ / Molar mass of KClO₃

Molar mass of KClO₃ = M_r(KClO₃) = 122.55 g/mol.

Finally, we can find the number of grams of KClO₃ that reacted:

m(KClO₃) = n(KClO₃) * M_r(KClO₃)

m(KClO₃) = (0.1716 moles) * (122.55 g/mol) = 21.0296 g = 21.0 g

m(KClO₃) = 21.0 g.

Answer: 21.0 grams of KClO₃ must have decomposed.