What volume of oxygen at 455 K and a pressure of 127400 Pa is produced by the decomposition of 114.7 g of BaO2 to BaO and O2?

10 L

Explanation:

The equation of the reaction is;

2BaO2 = 2BaO + O2

Number of moles of  BaO2 = 114.7 g/169.33 g/mol = 0.677 moles

From the reaction equation;

2 moles of BaO2 yields 1 moles of O2

0.677 moles of BaO2 yields 0.677 * 1/2 = 0.3385 moles of oxygen

Hence;

PV=nRT

V = ?

P = 127400 Pa or 1.257 atm

T = 455 K

n = 0.3385 moles

R = 0.082 atmLmol-1K-1

V = nRT/P

V = 0.3385 * 0.082 * 455/1.257

V= 10 L