cum
top of page

CSEC Chemistry: The Concept of the Limiting Reagent

In reactions that go to completion, all reactants are consumed given that they are mixed in the mole ratio shown by the balanced equation. For example:

2H₂(g) + O₂(g) → 2H₂O(l)

According to this equation, 2 mol of hydrogen gas reacts completely with 1 mol of oxygen gas to produce 2 mol of water. Or, if we use the molar masses of the substances, 4 g of hydrogen reacts completely with 32 g of oxygen to produce 36 g of water.

If we were to mix 7 g of hydrogen with 32 g of oxygen, then the hydrogen would be in excess. No more water would be produced once all the oxygen is used up, leaving 3 g of hydrogen unreacted. In the same way, if 4 g of hydrogen were mixed with 40 g of oxygen, the amount of water produced would be limited by the amount of hydrogen present, since the oxygen is present in excess. 8 g of oxygen would remain unreacted.

When two or more quantities of reactants are given, we must discern which reactant is the limiting reagent. The limiting reagent is the reactant which produces the smallest yield of products. In the first case, oxygen is the limiting reagent, while hydrogen is the limiting reagent in the second.


Example: If 20 g of Calcium carbonate is added to 1 dm³ of a solution containing 10 gdm⁻³ of hydrochloric acid, what mass of CO₂ is produced?


We can write the balanced equation first to find the ratio of reactants and products:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) +CO₂(g) +H₂0(l)


Then, we calculate the number of moles of each reactant:

mol of CaCO₃ = mass/molar mass = 20 g/100 gmol⁻¹ = 0.20 mol


mol of HCl = (10 gdm⁻³/36.5 mol) x 1 dm³ = 0.274 moldm⁻³ x 1 dm³= 0.274 mol HCl used


We then calculate the number of moles of product that can possibly be obtained from each number of moles of reactant.:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) +CO₂(g) +H₂0(l)

  • 0.2 mol CaCO₃ produces 0.20 mol CO₂

  • 0.274 mol HCl wil produce 0.137 mol CO₂

Remember that the reactant that produces the smaller amount of product is the limiting reagent. In this case, the limiting reagent is HCl. So, to find the mass of CO₂, we multiply the smaller number of moles of product by the molar mass of the product (CO₂):

0.137 mol x 44 gmol⁻¹ = 6.03 g


So, it is important that you are able to find the limiting reagent when given an arbitrary reaction between two or more reactants.

43 views0 comments

Recent Posts

See All

CSEC Chemistry: Complete Syllabus Review

Hello Chemistry students! This material was covered in Quelpr's CSEC Chemistry on June 26, 2021, however the past paper questions and solutions have been removed. If you are interested in similar cont

Comments


bottom of page