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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} HCl=(20g)/(36.5g/mol)=0.55moles`

`\text{Moles of} NaOH=(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=
`(1)/(1)* 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* {\text {Molar mass}}=0.14moles* 36.5g/mol=5.1g`

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