Balanced equation:
[tex]4H_3BO_3+2NaOH\rightarrow7H_2O_+Na_2B_4O_7[/tex][tex]\%yield=\frac{actual}{theoretical}\times100[/tex]We need to determine theoretically how many grams of sodium tetraborate can be produced from 45g of NaOH.
[tex]\begin{gathered} molNaOH=mass\div molecular\text{ }mass \\ NaOH\text{ }mol=45g\div40gmol^{-1} \\ NaOH\text{ }mol=1.13 \end{gathered}[/tex]Based on the mole ratio, 2 moles of NaOH produces 1 mole of sodium tetraborate.
[tex]\begin{gathered} mole\text{ }sodium\text{ }tetraborate=\frac{1.13}{2}=0.57 \\ mass\text{ }sodium\text{ }tetraborate=mole\times molecular\text{ }mass \\ mass\text{ }sodium\text{ }tetraborate=0.57mol\times381.37gmol^{-1} \\ mass\text{ }sodium\text{ }tetraborate=217.38g \end{gathered}[/tex]Theoretical value of sodium tetraborate is 217.38
[tex]\begin{gathered} \%\text{ }yield=\frac{93.5}{217.4}\times100 \\ \%\text{ }yield=43\% \end{gathered}[/tex]% yield is 43%
