Respuesta :
Answer: 27.5 liters
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For [tex]KClO_3[/tex]
Given mass = 100.0 g
Molar mass of [tex]KClO_3[/tex] = 122.5 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of}KClO_3 =\frac{100}{122.5}=0.82moles[/tex]
[tex]2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)[/tex]
2 moles of [tex]KClO_3[/tex] produces 3 moles of [tex]O_2[/tex]
0.82 moles of [tex]KClO_3[/tex] produces =[tex]\frac{3}{2}\times 0.82=1.23[/tex] moles of [tex]O_2[/tex]
Volume of [tex]O_2=moles\times {\text {molar volume}}=1.23\times 22.4=27.5L[/tex]
Thus 27.5 liters of [tex]O_2[/tex] are produced by the complete decomposition of 100.0 g of [tex]KClO_3[/tex] at STP
