Question

A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. Na + Cl2 → NaCl If the percentage yield of the reaction is 88%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.​

Answer 1

Answer: The actual yield of the product is 38.57 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of Na = 17.25 g

Molar mass of Na = 23 g/mol

Putting values in equation 1, we get:

$\text{Moles of Na}=\frac{17.25g}{23g/mol}=0.75mol$

For the given chemical reaction:

$2Na+Cl_2\rightarrow 2NaCl$

By stoichiometry of the reaction:

2 moles of Na produces 2 moles of NaCl

So, 0.75 moles of Na will produce = $\frac{2}{2}\times 0.75=0.75mol$ of NaCl

We know, molar mass of $NaCl$ = 58.44 g/mol

Putting values in above equation, we get:

$\text{Mass of NaCl}=(0.75mol\times 58.44g/mol)=43.83g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Percent yield of the product = 88 %

Theoretical value of the product = 43.83 g

Plugging values in equation 1:

$88\% =\frac{\text{Actual yield}}{43.83g}\times 100\\\\\text{Actual yield}=\frac{88\times 43.83}{100}=38.57g$

Hence, the actual yield of the product is 38.57 g