Limiting Reactants

• Reactions don't go on forever - you need stuff in the reaction flask that can react. If one reactant gets completely used up in a reaction before the reast, then the reaction will stop. That reactant's called limiting
• Reactions stop when one reactant is used up
• When some magnesium carbonate (MgCO3) is placed into a beaker of hydrochloric acid, you can tell a reaction is taking place because you see lots of bubbles of gas being given off. After a while, the amount of fizzing slows down and the reaction eventually stops
• 1)The reaction of magnesium carbonate (MgCO3) stops when all of one of the reactants is used up. Any other reactants are in excess

  2) The reactant that's used up in a reaction is called the limiting reactant (because it limits the amount of product that's formed)
• 3) The amount of product formed is directly proportional to the amount of limiting reactant. E.g. if you halve the amount of limiting reactant the amount of product formed will also halve. If you double the amount of limiting reactant the amount of product will double (as long as it's still the limiting reactant)

  4) This is because if you add more reactant there will be more reactant particles to take part in the reaction, which means more product particles
• The amount of product depends on the limiting reactant
• You can calculate the mass of a product formed in a reaction by using the mass of the limiting reactant and the balanced reaction equation:

  1)Write out the balanced equation

  2) Work out relative formula masses (Mr) of the reactant and product you want

  3) Use the balanced equation to work out how many moles there'll be of the other substance. In this case, that's how many moles of product will be made of this many moles of reactant

  4) Use the number of moles to calculate the mass
• Example

  Calculate the mass of aluminium oxide formed when 135 g of aluminium is burned in air

  1)Write out the balanced equation

  > 4Al + 302 -> 2Al2O3

  2) Calculate the relative formula masses

  > Al: 27 Al2O3: (2 x 27) + (3 x 16) = 102 (you don't have to find Mr of oxygen because it's in excess)

  3) Calculate the number of moles of aluminium in 135 g

  > Moles = mass / Mr = 135 / 27 = 5
• Example

  Calculate the mass of aluminium oxide formed when 135 g of aluminium is burned in air

  4) Look at the ratio of moles in the equatio

  > 4 moles of Al react to produce 2 moles of Al2O3 - half the number of moles are produced. So 5 moles of Al will react to produce 2.5 moles of Al2O3 (if the question asked for the number of moles of aluminium oxide formed, you'd stop here)

  5) Calculate the mass of 2.5 moles of aluminium oxide

  > Mass = moles x Mr = 2.5 x 102 = 255 g
• The mass of product is called the yield of a reaction. Masses you calculate in this way are called theoretical yields. In practice you never get 100% of the yield, so the amount of product you get will be less than you calculated