Consider the following reaction: 2K2O(s) + 2Cl2(g) -----4 KCl (s) + O2(g).

Question

If 5.00 g of K2O reacts completely with excess water:

a. What volume of O2 gas will be collected at STP?

b. If all the O2 gas is collected in a 0.500 L metal cylinder at 37.0˚C, what pressure will it exert?

Answer 1

2K₂O(s) + 2Cl₂(g) ==> 4KCl(s) + O2(g) ... balanced equation

(a) At STP, 1 mole of an ideal gas will occupy 22.4 L (you should memorize this fact). So, to find the volume of O₂ gas, all we need to do is find the moles of O₂ and multiply by 22.4 L/mole.

We find moles of O2 by using the stoichiometry of the balanced equation along with the molar mass of K2O (94.2 g/mol) and dimensional analysis.

moles O₂ produced = 5.00 g K₂O x 1 mol K₂O/94.2 g x 1 mol O₂ / 2 mol K₂O = 0.0265 moles O₂ produced

Volume O₂ = 0.0265 moles x 22.4 L / mole = 0.594 L

(b) To find the pressure under the described conditions, we can use the ideal gas law, PV = nRT

P = pressure = ? atmospheres

V = volume in L = 0.500 L

n = moles of gas = 0.0265 moles (calculated above)

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 37ºC + 273 = 310K

Solving for P we have:

P = nRT/V = (0.0265)(0.0821)(310) / 0.500

P = 1.35 atm