Answer: 8.41 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{3.29g}{32g/mol}=0.103moles[/tex]
The balanced chemical equation for decomposition is :
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
According to stoichiometry :
As 3 moles of [tex]O_2[/tex] are produced by = 2 moles of [tex]KClO_3[/tex]
Thus 0.103 moles of [tex]O_2[/tex] are produced by =[tex]\frac{2}{3}\times 0.103=0.0687moles[/tex] of [tex]KClO_3[/tex]
Mass of [tex]KClO_3=moles\times {\text {Molar mass}}=0.0687moles\times 122.5g/mol=8.41g[/tex]
8.41 g was the mass of the original sample.
