Question

100 gm of KMnO3 are heated to redness. what volume of ocygen at 39 C and 765 atm pressure will evolve ?

AnyOne .. explain please..

Answer 1

KMnO3 or KMnO4 ????

Answer 2

because if KMnO4 then
2 KMnO4 --Δ--> K2MnO4 + MnO2 + O2

Answer 3

Has to be KMnO4. There is no KMnO3

Answer 4

You have to set up the stoichiometry equation. Convert g of KMnO4 to moles of KMnO4. Then check the equation. And calculate the amount of Oxygen released with the 100 g (in moles) of KMnO4. Remember that 2 moles of KMnO4 gives you 1 mole of O2 gas. Then use PV=nRT and calculate the volume of oxygen evolved.

Answer 5

>>>> sorry
its KMnO4 ;)

Answer 6

om go for it

Answer 7

Can show me how to solve that please ?

Answer 8

moles of KMnO4 = mass/MM
n=100/(39+55+4*16)
n=100/158

2 KMnO4 --Δ--> K2MnO4 + MnO2 + O2
2 moles produses 1 mole
100/158 mole produses 50/158 mole

assume O2 ideal gas
P*V=n*R*T
V=n*R*T/P

n=50/158
R=0.082
T=(39+273)
P=765

Answer 9

thanks (: