Determine the mass of the salt produced when 325 ml of 0.35 M sulphuric acid is reacted with 458 ml of 0.222

Determine the mass of the salt produced when 325 ml of 0.35 M sulphuric acid is reacted with 458 ml of 0.222 M sodium hydroxide.

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

(0.325 L H2SO4) x (0.35 mol/L) = 0.11375 mol H2SO4
(0.458 L NaOH) x (0.222 mol/L) = 0.101676 mol NaOH

0.101676 mole of NaOH would react completely with 0.101676 x (1/2) = 0.050838 mole of H2SO4, but there is more H2SO4 present than that, so H2SO4 is in excess and NaOH is the limiting reactant.

(0.101676 mol NaOH) x (1/2) x (142.0428 g Na2SO4/mol) = 7.22 g Na2SO4