Answer to Question #124061 in General Chemistry for K

Question

A 34.2 gram sample of impure potassium nitrate (KNO3) was heated to complete decomposition according to the equation
2 KNO3(s) → 2 KNO2(s) + O2(g)
After the reaction was complete, the volume
of the dry gas produced was 1 liters at 75◦C
and 721 torr. What was the percent KNO3
(101.1 g/mol) in the original sample? (Assume that only the potassium nitrate had
decomposed.)
Answer in units of %.

Answer 1

pV=nRT

n=pV/RT = 96.1254 x 1 / 8.314 x 348 =

= 0,0332237687 moles of O2.

0,0332237687 x 2 x 101.1 = 6,71784603 g of KNO3.

6,71784603 x 100% / 34.2 = 19,6428246 %.