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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 goes to completion, how much of which reactant is left in excess?

User SytS
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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}=(26.5g)/(53.5g/mole)=0.495moles


\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=(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* \text{Molar mass of }NH_4Cl


\text{Mass of }NH_4Cl=(0.245mole)* (53.5g/mole)=13.1075g

Therefore, the amount of reactant left in excess is 13.1075 grams.

User Divyesh Gondaliya
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