Answer: 5.1 gram
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} HCl=\frac{20g}{36.5g/mol}=0.55moles[/tex]
[tex]\text{Moles of} NaOH=\frac{16.3g}{40g/mol}=0.41moles[/tex]
[tex]HCl(aq)+NaOH(s)\rightarrow NaCl(aq)+H_2O(l)[/tex]
According to stoichiometry :
1 mole of [tex]NaOH[/tex] require = 1 mole of
Thus 0.41 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.41=0.41moles[/tex] of
Thus [tex]NaOH[/tex] is the limiting reagent as it limits the formation of product and [tex]HCL[/tex] is the excess reagent.
Moles of HCl left = (0.55-0.41) = 0.14
Mass of [tex]HCl[/tex] left =[tex]moles\times {\text {Molar mass}}=0.14moles\times 36.5g/mol=5.1g[/tex]
Thus 5.1 g of hydrochloric acid could be left over by the chemical reaction.
