Answer
(a)Theoretical yield of $O_2$=2.26g
(b)Percent yeild=46.46%
Work Step by Step
$KClO_3$ molar mass=122.55amu
$O_2$ molar mass=32amu
(a)We calculate the number of moles of $KClO_3$ using the formula:
Number of moles=$\frac{\text{mass}}{\text{molecular mass}}=\frac{5.79}{122.55}\approx0.047mol$
We write a balanced chemical reaction for the decomposition of $KClO_3$ to potassium chloride and water:
$2KClO_3\rightarrow 2KCl+3O_2$
According to the balanced chemical reaction, 2moles of $KClO_3$ will produce 3 moles of $O_2$. So, 0.047mol $KClO_3$ will produce:
$0.047mol $KClO_3$\times\frac{3mol(O_2)}{2mol($KClO_3$)}=0.0705mol$
We can then calculate the mass of oxygen using the formula:
mass of oxygen=number of moles of oxygen$\times$molar mass of oxygen($O_2$)=$0.0705\times32\approx2.26g$
Theoretical yield of oxygen gas=2.26g
(b)Percent yeild=$\frac{\text{Actual yeild}}{\text{theoritical yeild}}\times100=\frac{1.05}{2.26}\times100\approx46.46\%$
