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

##### 2 Answers

Dec 21, 2017

Look at the stoichiometric equation....

#### Explanation:

A little

We require

And so we need

Dec 21, 2017

#### Explanation:

- Write and balance the equation

#2KClO_3->2KCl+3O_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)# - 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# - 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# - 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# - Therefore,
#5.5gO_2# is produced in the decomposition of#14.04gKClO_3# .