According to this reaction, the ratio between Al(CN)3 and H2SO4 is 2:3. Knowing the molarity of H2SO4, in order to find out the volume we need to find out the moles of H2SO4 first.
Molar mass of Al(CN)3 is 26.98+(12.01+14.01)*3 = 105.04 g/mol. So the moles of Al(CN)3 reacted = mass/molar mass = 0.4198g/105.04g/mol = 0.003997mol. According to the above ratio, the moles of H2SO4 = 0.003997mol*3/2 = 0.005995mole.
The volume of H2SO4 needed for the reaction = moles of H2SO4/molarity of H2SO4 = 0.005995mole/0.3682mol/L = 0.01628L = 16.28 mL
