Question

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 1.5 g of hydrochloric acid is mixed with 2.67 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer: The maximum amount of water that can be produced is 0.74 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)  

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid:}=\frac{1.5g}{36.5g/mol}=0.041mol$

$\text{Moles of sodium hydroxide}=\frac{2.67g}{40g/mol}=0.067mol$

The chemical equation for the reaction is

$HCl+NaOH\rightarrow NaCl+H_2O$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of NaOH

So, 0.041 moles of HCl will react with = $\frac{1}{1}\times 0.041=0.041mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent. Thus, HCl is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of HCl produces = 1 mole of water

So, 0.041 moles of HCl will produce = $\frac{1}{1}\times 0.041=0.041moles$ of water

Mass of water=$moles\times {\text{Molar Mass}}=0.041mol\times 18g/mol=0.74g$

Thus the maximum amount of water that can be produced is 0.74 g