Answer
$97.8\%$
Work Step by Step
$Sb_{4}O_{6}$ is the limiting reactant. The theoretical yield of $Sb$ is the mass of $Sb$ produced based on the amount of $Sb_{4}O_{6}$.
$60.0\,g\times\frac{1\,mol\,Sb_{4}O_{6}}{583.0364\,g}\times\frac{4\,mol\,Sb}{1\,mol\,Sb_{4}O_{6}}\times\frac{121.76\,g}{1\,mol\,Sb}=50.121\,g$
$\text{% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%=\frac{49.0\,g}{50.121\,g}\times100\%=97.8\%$
