Calculate the volume of concentrated HCl (12 M) needed to convert the sodium benzoate back to benzoic acid.

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.

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

3 hours because it is a decent distance.