Answer: 12 grams of oxygen
Explanation:
To calculate the moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\)
For \(KClO_3\)
Given mass = 31.9 g
Molar mass of \(KClO_3\) = 122.5 g/mol
Putting values in above equation, we get:
\(\text{Moles of}KClO_3 =\frac{31.9}{122.5}=0.26moles\)
\(2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)\)
2 moles of \(KClO_3\) produces 3 moles of \(O_2\)
0.26 moles of \(KClO_3\) produces =\(\frac{3}{2}\times 0.26=0.39\) moles of \(O_2\)
Mass of \(O_2=moles\times {\text {molar mass}}=0.39mol\times 32g/mol=12g\)
Thus 12 grams of oxygen was produced.
