The question is incomplete, the complete question is;
Calculate the volume of concentrated HCL (12 M) needed to convert 1.5 g of sodium benzoate back to benzoic acid.
Answer:
8.33 * 10^-4 L
Explanation:
Equation of the reaction;
C6H5COONa + HCl -------->>>> C6H5COOH + NaCl
Number of moles of sodium benzoate = mass/molar mass
molar mass of sodium benzoate =144.11 g/mol
mass of sodium benzoate = 1.5 g
Number of moles of sodium benzoate = 1.5g/144.11 g/mol
Number of moles of sodium benzoate = 0.01 moles
Since the reaction is 1:1, 0.01 moles of HCl reacted.
but;
n = CV
n = number of moles
C = concentration
V = volume
V = n/C
V = 0.01 moles/12 M
V = 8.33 * 10^-4 L
