What volume of oxygen gas?

Question

Is produced at 751 torr and 23*C when 6.48 grams of KCLO*3 is decomposed by the following reaction: 2KCLO*3 --> KCL (s) + O*2 (g)

* means subscripts

Answer 1

First find out how many moles of oxygen gas are produced from the balanced chemical equation. Then use the ideal gas equation pv = nRT and solve for v, the volume of O2 (g).

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

The molar mass of KClO3 is 122.55 g/mol

(6.48 g KClO3)( 1 mol KClO3/ 122. 55 g KClO3)(2 mol KClO3/ 3 mol O2) = 0.0353 mol O2

pv = nRT where R, the idea gas constant is 0.08206 (L*atm/mol*K)

solve for v of O2:

v = (nRT)/(p)

we have to convert the pressure to atm:

(751 torr)(1 atm/760 torr) = 0.988 atm

also convert the temperature to Kelvins:

23 + 273.15 K = 296.15 K

v = [(0.0353 mol)(0.08206 L*atm/mol*K)(296.15 K)]/(0.988 atm) = 0.868L

Hope I did that correctly. ;)