If we react 1g of Ca with 1g of Cl2, how many grams of CaCl2 will be formed as per the following equation?
Ca + Cl2 → CaCl2
Using the Law of Conservation of Mass we can say that the total weight of the reactants is 2g and hence the weight of the product (CaCl2) should also be 2g.
According to the Law of Conservation of Mass, the mass of the reactants should be equal to the mass of the products if all the reactants are consumed in the reaction. The catch is we will be able to us the Law of Conservation of Mass only if all the reactants take part in the reaction.
What if all the reactants are not consumed in the reaction? This brings in the concept of “Limiting Reagent”.
Limiting Reagent in a reaction is that reactant that limits the amount of product that can be formed. The reaction will not proceed forward when this reactant is consumed completely. The other reactants may be left unreacted in the reaction mixture.
Let us take the example of an imaginary bicycle company to understand this. A bicycle has many parts. Some of them are the two wheels, the frame, the pedals, the tires, the handle bars, the brake shoes and the seat. This bicycle company has many independent units manufacturing these different parts that are needed to assemble one complete bicycle. Say, the wheel unit makes 1000 wheels in a day, the frame unit makes 1200 frames a day, the tire unit makes 1000 tires a day, the handle bar unit makes 2000 a day, the brake shoe unit makes 5000 a day and the seat unit makes 1 seat a day. How many complete bicycles can this company send out in a day? The answer is 1. Irrespective of the number of other parts manufactured, the cycle will not be complete without the seat, which is manufactured 1 a day.
Similarly, that reactant that brings less than required number of moles (or mass) for a reaction will be the limiting reagent.
Let us now get back to our question. According to the balanced equation, 1 mole of Ca will react with 1 mole of Cl2 and give 1 mole of CaCl2.
We have 1g of Ca which is (1⁄40) mole and 1g of Cl2 which is (1⁄71) mole.
(1⁄71) is less than (1⁄40) numerically.
This means that there is not enough Cl2 molecules to react with all the Ca atoms present. Hence, Cl2 is the “Limiting Reagent” in this reaction.
Now that we have identified the Limiting Reagent in this reaction, we will now base all our calculations on the weight or the number of moles of the Limiting Reagent.
According to the balanced equation, 1 mole of Cl2 gives 1 mole of CaCl2.
So, (1⁄71) mole of Cl2 will produce (1⁄71) mole of CaCl2.
The weight of CaCl2 produced thus will be (1⁄71 * 111) = 1.56g
Also, (1⁄71) mole of Cl2 will react only with (1⁄71) mole of Ca.
So, the mole of Ca left unreacted will be (1⁄40) - (1⁄71) = 0.011 mole.
Hence the weight the Ca left unreacted will be (0.011 * 40) = 0.44g
We will now be left with 1.56g of CaCl2 and 0.44g of unreacted Ca, the sum of which is the 2g of reactants taken.
So, if any of the reactant is left unreacted in a reaction, the total of the weight of the products formed and the weight of the reactant left unreacted will be equal to the weight of the reactants taken at the beginning for the reaction.
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The Law of Conservation of Mass
In simple words the weight of the reactants are always equal to the weight of the products if all reactants are consumed in the reaction."Mass can neither be created nor destroyed"
Law of Conservation of Mass was introduced by Antoine Lavoisier in around 1789. He had conducted many experiments especially of combustion reactions in closed containers to validate the rule.
In Stoichiometry we use the Law of conservation of mass to calculate various quantities, be it the amount of product expected or the amount of reactant needed to be added to a reaction or the percent yield of a reaction.
Let us take an example;
CaCO3 on thermal decomposition produces CaO and CO2.
CaCO3 → CaO + CO2
On complete decomposition of 100g of CaCO3 produced 56g of CaO and 44g of CO2 The total weight of the reactants is equal to the total weight of the products.
A 100g sample of CaCO3 on thermal decomposition gives 40g of CaO. What can we infer from this?
We can infer either of the following two things.
1. The reaction is incomplete and
2. The sample is impure.
Let us say that the reaction is complete. This leaves us with the only option that the sample must have been impure. What must be the percentage purity of the sample? Let us calculate.
We know that, 100g of pure CaCO3 on complete decomposition should produce 56g of CaO. Or in other words we can say that 56g of CaO should have been produced from 100g of pure CaCO3. So, 40g of CaO must have been produced from [(100/56) * 40]g = 71.43g of CaCO3. Hence the sample must be 71.43% pure.
What if the reactants did not completely react? How will we be able to calculate the amount of products formed and also the amount of reactants reacted?
The above questions can be answered by using the concept of “Limiting Reagent”. We shall discuss about it next.
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