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

That result from subduction of oceanic plates beneath lighter continental plates. most of the earth's volcanoes are located around the pacific ring of fire because that the location of most of the earth's subduction zones.