What mass of #"potassium chlorate"# is necessary to generate a #5.5*g# mass of oxygen gas?

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

A little #Mn(+IV)# salt is usually added to catalyze the reaction....

We require #5.5*g# dioxygen gas....a molar quantity of #(5.5*g)/(32.0*g*mol^-1)=0.0172*mol#, and note that this is in respect to dioxygen gas....

And so we need #2/3*"equiv"# with respect to the chlorate....i.e. #2/3xx0.0172*molxx122.55*g*mol^-1=14.0*g#

1. Write and balance the equation
   #2KClO_3->2KCl+3O_2#
2. First thing to do is to find the molar masses of the involved compounds. In this case, #O_2# and #KClO_3#. Refer to the periodic table for the elements' atomic masses.
   #O_2=(32g)/(mol)#
   #KClO_3=(122.5g)/(mol)#
3. Given the mass of #O_2#, as convention, convert #"mass"(m)# to #"moles"(eta)# as basis for the usual series of conversions.
   #=5.5cancel(gO_2)xx(1molO_2)/(32cancel(gO_2))#
   #=0.1719molO_2#
4. Now, since the target is the #mKCl)_3#, use #etaO_2# as basis with reference to the balanced equation for the mole ratio to find the #etaKClO_3#;i.e.,
   #=0.1719cancel(molO_2)xx(2molKClO_3)/(3cancel(molO_2))#
   #=0.1146molKClO_3#
5. Finally, find #mKClO_3# through the relationship obtainable from the molar mass of #KClO_3#; that is, #1molKClO_3-=122.5gKClO_3#. Hence,
   #=0.1146cancel(molKClO_3)xx(122.5gKClO_3)/(1cancel(molKClO_3))#
   #=14.04gKClO_3#
6. Therefore, #5.5gO_2# is produced in the decomposition of #14.04gKClO_3#.