Question

Please balanceH3BO3 + NaOH ----> H2O + Na2B4O7If you start with 45 grams of NaOH, you produce 93.5 grams of sodium tetraborate. What is your percent yield?

Answer 1

Balanced equation:

`4H_3BO_3+2NaOH\rightarrow7H_2O_+Na_2B_4O_7`
`\%yield=(actual)/(theoretical)*100`

We need to determine theoretically how many grams of sodium tetraborate can be produced from 45g of NaOH.

`\begin{gathered} molNaOH=mass/ molecular\text{ }mass \\ NaOH\text{ }mol=45g/40gmol^(-1) \\ NaOH\text{ }mol=1.13 \end{gathered}`

Based on the mole ratio, 2 moles of NaOH produces 1 mole of sodium tetraborate.

`\begin{gathered} mole\text{ }sodium\text{ }tetraborate=(1.13)/(2)=0.57 \\ mass\text{ }sodium\text{ }tetraborate=mole* molecular\text{ }mass \\ mass\text{ }sodium\text{ }tetraborate=0.57mol*381.37gmol^(-1) \\ mass\text{ }sodium\text{ }tetraborate=217.38g \end{gathered}`

Theoretical value of sodium tetraborate is 217.38

`\begin{gathered} \%\text{ }yield=(93.5)/(217.4)*100 \\ \%\text{ }yield=43\% \end{gathered}`

% yield is 43%