Question: How can I find limiting reagent with moles?...

Question

How can I find limiting reagent with moles?

Answer 1

Solution

In a reaction, there are two types of reactants: excess reactant and limiting reactant, the formation of the product is calculated by the amount of the limiting reactant. First, balance the equation, and then find the number of moles, those which will form a lesser amount of products will be limiting reactants.

Complete answer:
In a reaction, there are two types of reactants: excess reactant and limiting reactant, the formation of the product is calculated by the amount of the limiting reactant.
For finding the limiting reagents, first, you have to write the chemical reaction, then balance the reaction correctly. Then with the help of a given mass of the reactants, find the number of moles of the reactants. Now, with the help of moles of reactants to calculate the mass of products formed in the reaction, those which will form a lesser amount of products will be limiting reactants.
For example, when we take 0.25 grams of NaCl and 0.25 grams of $AgN{{O}_{3}}$ are reacted then AgCl and $NaN{{O}_{3}}$ are formed, and AgCl is formed as precipitate and will decide the amount of the formation of AgCl.
We can write the reaction as:
$NaCl+AgN{{O}_{3}}\to AgCl+NaN{{O}_{3}}$
This reaction is balanced, so we see that 1 mole of both sodium chloride and silver nitrate reacts to form one mole of silver chloride.
The given mass of sodium chloride is 0.25, so the number of moles will be:
$\dfrac{0.25}{58.5}=0.0042$
Therefore, 0.0042 moles of AgCl will be formed and it is equal to 0.61 grams.
The given mass of silver nitrate is 0.25, so the number of moles will be:
$\dfrac{0.25}{169.87}=0.0014$
Therefore, 0.0014 moles of AgCl will be formed and it is equal to 0.21 grams.
So, here we can see that silver nitrate is the limiting reagent because it forms a lesser amount of silver chloride.

Note:
If the reaction is not balanced, then you cannot find the correct limiting and excess reagent. To convert the moles into grams multiply the given moles with the molecular mass of the compound.