Question

Ill give the brainliest answer to whoever helps me with this equation

Answer 1

Answer: The percent yield for the $NaBr$ is, 86.7 %

Explanation : Given,

Moles of $FeBr_3$ = 2.36 mol

Moles of $NaBr$ = 6.14 mol

First we have to calculate the moles of $NaBr$

The balanced chemical equation is:

$2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr$

From the reaction, we conclude that

As, 2 moles of $FeBr_3$ react to give 6 moles of $NaBr$

So, 2.36 moles of $FeBr_3$ react to give $\frac{6}{2}\times 2.36=7.08$ mole of $NaBr$

Now we have to calculate the percent yield for the $NaBr$.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%$

Therefore, the percent yield for the $NaBr$ is, 86.7 %