Answer: 0 grams as HBr is the limiting reagent and gets completely consumed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} HBr=\frac{17.0g}{81g/mol}=0.210moles[/tex]
[tex]\text{Moles of} NaOH=\frac{14.0g}{40g/mol}=0.350moles[/tex]
[tex]HBr+NaOH\rightarrow NaBr+H_2O[/tex]
According to stoichiometry :
1 mole of [tex]HBr[/tex] require = 1 mole of [tex]NaOH[/tex]
Thus 0.210 moles of [tex]HBr[/tex] will require=[tex]\frac{1}{1}\times 0.210=0.210moles[/tex] of [tex]NaOH[/tex]
Thus [tex]HBr[/tex] is the limiting reagent as it limits the formation of product and [tex]NaOH[/tex] is the excess reagent.
Thus no mass of hydrobromic acid that could be left over by the chemical reaction
