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Chemistry

1 answer:

Nana76 [90]3 years ago

3 0

First on is transport

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A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go

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Answer : The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of $NH_4Cl$ = 26.5 g

Mass of $NaOH$ = 10 g

Molar mass of $NH_4Cl$ = 53.5 g/mole

Molar mass of $NaOH$ = 40 g/mole

First we have to calculate the moles of $NH_4Cl$ and $NaOH$ .

$\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NH_4Cl+NaOH\rightarrow NH_4OH+NaCl$

From the balanced reaction we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $NH_4Cl$

So, 0.25 moles of $NaOH$ react with 0.25 moles of $NH_4Cl$

From this we conclude that, $NH_4Cl$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of $NH_4Cl$ .

$\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl$