Question

If 283g KClO3 react, how many grams of O2 will be produced?

Answer 1

`110.88 \ "g"` of oxygen is produced.

Explanation:

First, we must write the balanced equation for the decomposition of potassium chlorate:

`2KClO_3rarr2KCl+3O_2`

Note that the potassium chlorate and the oxygen are in the ratio `2:3`. That is to say, for every `2` moles of potassium chlorate which react, `3` moles of oxygen are produced.

The molar mass of potassium chlorate is `122.55"g/mol"`.

The formula for moles, `n`, is `n=m/M`, where `m` is the given mass and `M` the molar mass.

`n=283/122.55~~2.31"mol"` of potassium chlorate.

Remember that for every `2` moles of potassium chlorate which react, `3` moles of oxygen are produced.

So when `x` moles of oxygen are produced,
`2/3=2.31/x`

`2x=3*2.31`

`x~~3.47"mol"` of oxygen is produced.

You can rearrange the formula for moles to solve for `m`:

`n=m/M`

`m=nM`

The molar mass of oxygen is `32"g/mol"`, so we can input:

`m=32*3.47`

`m=110.88 \ "g"` of oxygen is produced.