Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. What is the theoretical yield of water formed from the reaction of 1.1g of hydrochloric acid and 2.1g of sodium hydroxide?

Answer 1

Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

$\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

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

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $HCl$ react to give 1 mole of $H_2O$

So, 0.030 moles of $HCl$ react to give 0.030 moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

$\text{ Mass of }H_2O=(0.030moles)\times (18g/mole)=0.54g$

Therefore, the theoretical yield of water formed from the reaction is 0.54 grams.