Which of these is NOT found in the earths crust?OxygenIronSulphurHelium

Ответ:

iliketurtures

13.10.2020

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%

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