Question

Determine the percent yield forthe reaction between 3.74 g of Naand excess O2 if 5.34 g of Na2Ois recovered.

Answer 1

Answer

105.95%

Explanation

Given:

Mass of Na = 3.74 g

Mass of Na2O recovered (actual yield) = 5.34 g

What to find:

The percent yield for the reaction

Step-by-step solution:

The first step is to write a balanced chemical equation for the reaction.

$4Na+2O_2\rightarrow2Na_2O$

The next step is to calculate the theoretical yield of Na2O.

From the Periodic Table:

Molar mass of Na = 22.99 g/mol

Molar mass of Na2O = 61.98 g/mol

From the equation above, 4 moles of Na produced 2 moles of Na2O

In grams, 4 x 22.99 = 91.96 g of Na produced x 61.98 = 123.96 g of Na2O

So 3.74 g of Na will produce:

$\frac{3.74g\text{ }Na\times123.96\text{ }g\text{ }Na_2O}{91.96\text{ }g\text{ }Na}=5.04\text{ }g\text{ }Na_2O$

The theoretical yield of Na2O = 5.04 g

Therefore, the percent yield can of the reaction can be calculated as follows:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{5.34\text{ }g}{5.04\text{ }g}\times100\% \\ \\ Percent\text{ }yield=1.0595\times100\% \\ \\ Percent\text{ }yield=105.95\% \end{gathered}$