What mass of #"potassium chlorate"# is required to produce a #2.24*L# volume of dioxygen gas at #"NTP"#?

1 Answer
Aug 7, 2017

Answer:

We need (i) a stoichiometric equation, and (ii)......

Explanation:

and........(ii) we need the molar volume at #NTP#, which I think is #24.04*L*mol^-1#.......

Now as regards (i)......

#KClO_3(s) + Delta stackrel(MnO_2)rarrKCl(s) + 3/2O_2(g)uarr#

The reaction generally works well if you add some #"manganese (IV) oxide"# as an oxygen transfer reagent.

And so moles of dioxygen gas......#=(2.24*L)/(24.04*L*mol^-1)#

#=0.0932*mol#

And so the molar quantity of #KClO_3# is......

#2/3xx0.0932*molxx122.55*g*mol^-1=7.61*g#.

Look at the question again, are you sure that it did not ask for #"STP"#? The different definitions and standards vary across each syllabus; there are different definitions in yurope, Blighty, and the States. So you need to look up your definition of #"NTP"#.