Answer:
1 mole KClO3 / 122.55 g KClO3,
3 mole O2 / 2 mole KClO3,
31.998 g O2 / 1 mole O2.
Explanation:
First, let's convert 10 g of KClO3 to moles using the molar mass of KClO3 which is 122.5 g/mol:
[tex]10\text{ g KClO}_3\cdot\frac{1\text{ mol KClO}_3}{122.5\text{ g KClO}_3}=0.082\text{ moles KClO}_3.[/tex]
Now you can see in the chemical equation that 2 moles of KClO3 reacted produces 3 moles of O2, so we state a rule of three to find the number of moles of O2:
[tex]\begin{gathered} 2\text{ mole KClO}_3\rightarrow3\text{ mole O}_2 \\ 0.082\text{ moles KClO}_3\rightarrow?\text{ mole O}_2 \end{gathered}[/tex]
The calculation of this will look like this:
[tex]0.082\text{ mole KClO}_3\cdot\frac{3\text{ mole O}_2}{2\text{ mole KClO}_3}=0.12\text{3 moles O}_2.[/tex]
The final step to finding the mass of oxygen (O2) is using its molar mass which is 32 g/mol, like this:
[tex]0.123\text{ mole O}_2\cdot\frac{32\text{ g O}_2}{1\text{ mol O}_2}=3.94\text{ g O}_2.[/tex]
As you can see from this process, the needed ratios are:
1 mole KClO3 / 122.55 g KClO3,
3 mole O2 / 2 mole KClO3,
31.998 g O2 / 1 mole O2. (in the explanation is approximated)
