Limiting Reagent and Calculating amount of EXCESS reactant.
1) Try the “light blue example #2 below. READ my notes on the side.
2) Try the two problems below the blue box.
In this example, the Mg is the limiting reactant, therefore, the oxygen is in excess, meaning the 10 grams
they provided you with was MORE than was needed to react with the 2.4 grams of the magnesium. In
order to determine how much oxygen was leftover at the time the reax stopped, you would want to
know the least amount of oxygen they needed to give you to react with all the Mg. There are two ways
to determine that minimum amount:
Determine the relationship between the LR and the oxygen. According to the molar
relationships from the balanced equation, for every 2 moles of Mg that are used, only one
mole of oxygen is necessary. So get out of grams of Mg to moles of Mg, then you can
compare moles of Mg to moles of oxygen. Lastly you can change your moles of oxygen to
grams of oxygen. This # of grams of oxygen (1.58 g) is all that was needed to be given to
react with all the Mg. BUT they gave you 10 grams of oxygen to play with. Therefore 10g1.58 g = 8.42 grams of oxygen that weren’t necessary, or which were leftover, in excess.
See letter (a) in the blue box above.
Determine the relationship between the amount of product that was made and how much
oxygen was necessary to make that amount of product. See letter (b) in the blue box above.
SHOW ALL WORK FOR THE PRACTICE PROBLEMS BELOW
Problem #1
If 298.15 g silicon reacts with 9.89 moles of nitrogen, how many moles of the product trisilicon
tetranitride are formed? Which if any reagent limited this reaction? What amount of excess reagent
remains?
Problem #2
Cupric chloride reacts with potassium iodide to produce potassium chloride, cuprous iodide, and
elemental iodine. If 68.45 g of cupric chloride react with 72.11 g of potassium iodide, what mass of
elemental iodine is formed? What reactant limits this reaction? How much reactant remains “in
excess?” What mass of potassium chloride forms?