What volume of 0.3682 m h2so4 solution is required to react with 0.4198 grams of al(cn)3 according to the reaction, 2 al(cn)3 + 3 h2so4  6 hcn + al2(so4)3?

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