Question

Aqueous hydrobromic acid HBr reacts with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O . If 0.555g of sodium bromide is produced from the reaction of of 2.4g hydrobromic acid and 0.77g of sodium hydroxide, calculate the percent yield of sodium bromide.

Answer 1

Answer: 28 %

Explanation:

First we have to calculate the moles of $NaOH$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{0.77g}{40g/mole}=0.019moles$

Now we have to calculate the moles of $HBr$

$\text{Moles of }HBr=\frac{\text{Mass of }HBr}{\text{Molar mass of }HBr}=\frac{2.4g}{81g/mol}=0.030moles$

The balanced chemical reaction will be,

$NaOH(aq)+HBr(aq)\rightarrow NaBr(aq)+H_2O(l)$

From the balanced reaction, we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $HBr$

So, 0.019 moles of $NaOH$ react with= $\frac{1}{1}\times 0.019=0.019$ mole of $HBr$

Thus $NaOH$ acts as the limiting reagent as it limits the formation of product and [tex}HBr[/tex] is the excess reagent.

As, 1 mole of $NaOH$ give= 1 mole of $NaBr$

So, 0.019 moles of $NaOH$ react with= $\frac{1}{1}\times 0.019=0.019$ mole of $NaBr$

Now we have to calculate the mass of $NaBr$

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(0.019mole)\times (103g/mole)=2.0g$

Now we have to calculate the percent yield of $NaBr$

$\%\text{ yield of }NaBr=\frac{\text{Actual yield of }NaBr}{\text{Theoretical yield of }NaBr}\times 100=\frac{0.555g}{2.0g}\times 100=27.75\%=28\%$

Therefore, the percent yield of sodium bromide is 28%