To grow alum crystals you decided to prepare your own aluminum sulfate. You have 23.4 g of aluminum filings and bottle of 2.00 M sulfuric acid. What volume of acid is needed to dissolve the aluminum?

As usual, we will start with balanced reaction equation:

2Al + 3H_{2}SO_{4} → Al_{2}(SO_{4})_{3} + 3H_{2}

However, so far we have used it always to calculate mass of reagent using known mass of other reagent. While we can do the same this time, what we really need is information about number of moles of sulfuric acid - as number of moles will be much easier to convert to volume of solution with known molarity. So instead of calcuating mass of needed sulfuric acid, we will set up our proportion in such a way that it will contain moles and mass at the same time:

2Al | + | 3H_{2}SO_{4} | → | Al_{2}(SO_{4})_{3} | + | 3H_{2} |

53.96 g | : | 3 moles | ||||

23.4 g | : | x moles |

Proportion looks like

53.96 g : 3 moles = 23.4 g : x mole

and x = 23.4 g/ 53.96 g × 3 mole = 1.30 mole

So you need 1.30 mole of sulfuric acid. What volume will it be? We can take molar concentration definition, solve it for V and insert known numbers into the equation:

V = n/C = 1.30/2.00 = **0.650 L** (or 650 mL)

And that's the final answer. In the real world you will want to use some excess of acid, it will be always worth to check if the solution is not oversaturated once the reaction ends - but that's completely different story.