How many grams of potassium chlorate are required in the preparation of 90.0 cubic decimeters of oxygen?

Question

I know the balanced equation is #2KClO_3 -> 3O_2 + 2KCl#.

Kind of confused b/c we convert from a gas (oxygen at 22.4 L/mol STP) to potassium chlorate (solid)

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Answer 1 · 2018-01-14T08:48:55

I make it approx...#300*g# #"chlorate salt...."#

You gots....

#KClO_3(s) +Deltastackrel{Mn^(+IV)}rarrKCl(s) + 3/2O_2(g)#

We want #90*dm^3-=90*L# dioxygen gas...

#"Moles of dioxygen"=(90*L)/(22.4*L*mol^-1)=4.02*mol#

And given this requirement we need AT LEAST....

#4.02*molxx2/3xx122.55*g*mol^-1=??*g# with respect to chlorate salt....

Note that especially in the UK we use #1*dm^3-=1*L#...How?

Well................,

#1*dm^3=(1xx10^-1*m)^3=1xx10^-3*m^3=1/1000*m^3#

#-=1*L#...cos there are #1000*L*m^-3#....