Answer: The actual yield of [tex]B_2O_3[/tex] is 60.0 g
Explanation:-
The balanced chemical reaction :
[tex]B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)[/tex]
Mass of [tex]B_2H_6[/tex] =[tex]Density\times Volume=1.131g/ml\times 36.9ml=41.7g[/tex]
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles[/tex]
According to stoichiometry:
1 mole of [tex]B_2H_6[/tex] gives = 1 mole of [tex]B_2O_3[/tex]
1.51 moles of [tex]B_2H_6[/tex] gives =[tex]\frac{1}{1}\times 1.51=1.51[/tex] moles of [tex]B_2O_3[/tex]
Theoretical yield of [tex]B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g[/tex]
Percent yield of [tex]B_2O_3[/tex]= [tex]75.7\%[/tex]
[tex]\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]75.7\%=\frac{\text{Actual yield}}{79.3}\times 100[/tex]
[tex]{\text{Actual yield}}=60.0g[/tex]
Thus the actual yield of [tex]B_2O_3[/tex] is 60.0 g
