Question

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

Answer 1

Look at the stoichiometric equation....

Explanation:

`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`

Answer 2

`=14.04gKClO_3`

Explanation:

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`.