For the reaction,
#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#........
Clearly, there is #1:1# stoichiometry.
#"Moles of HCl"=(7.81*g)/(36.5*g*mol^-1)=0.214*mol.#
#"Moles of NaOH"=(5.24*g)/(40.0*g*mol^-1)=0.131*mol.#
Clearly, #HCl# is in excess. Note that #HCl# is a room temperature gas (that has some great solubility in water). Normally, an aqueous solution of a known concentration would be proposed for the acid, and you would access the molar quantity of #HCl# by the relation:
#"Concentration"="Moles of HCl"/"Vollume of solution"#
