Question

g Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 20. g of hydrochloric acid is mixed with 16.3 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: 5.1 gram

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} HCl=\frac{20g}{36.5g/mol}=0.55moles$

$\text{Moles of} NaOH=\frac{16.3g}{40g/mol}=0.41moles$

$HCl(aq)+NaOH(s)\rightarrow NaCl(aq)+H_2O(l)$

According to stoichiometry :

1 mole of $NaOH$ require = 1 mole of

Thus 0.41 moles of $NaOH$ will require=$\frac{1}{1}\times 0.41=0.41moles$  of

Thus $NaOH$ is the limiting reagent as it limits the formation of product and $HCL$ is the excess reagent.

Moles of HCl left = (0.55-0.41) = 0.14

Mass of $HCl$ left =$moles\times {\text {Molar mass}}=0.14moles\times 36.5g/mol=5.1g$

Thus 5.1 g of hydrochloric acid  could be left over by the chemical reaction.