Question

How can we identify the limiting reactant when $43.25g$ of $Ca{{C}_{2}}$ reacts with $33.71g$ of water to produce $Ca\left( OH \right)$ to ${{C}_{2}}{{H}_{2}}?$

Answer 1

In a chemical reaction, there are two types of reagents present.one is excess reagent which is not consumed completely in the reaction and another one is consumed and the product is limited by that reagent. Limiting reagent: The reactant in a chemical reaction which is completely consumed when the reaction is completed. How much product will be formed is determined by the limiting reagent.

Complete answer:
We can find out the limiting reagent by a trick or you can say a formula that is whose ratio of mole to stoichiometric coefficient is low is known as limiting reagent. Some rules should be followed to find out the limiting reagent;
-Balance the equation.
-Convert it into the form of mole.
-Calculate the ratio of mole to stoichiometric coefficient.
-Calculate the mass of the product.
You can calculate the remaining excess reagent also.
$Ca{{C}_{2}}+2{{H}_{2}}O\to {{C}_{2}}{{H}_{2}}+Ca{{(OH)}_{2}}$
Notice that we have a $1:2~$ mole ratio between $Ca{{C}_{2}}~$ and ${{H}_{2}}O$ ; that is, every mole of the former used requires two moles of the latter. Since the quantities of both reactants are given, and knowing that their molar masses are $62g/mol(Ca{{C}_{2}})$ and $18g/mol({{H}_{2}}O)$ we can determine the number of moles from;
${{n}_{{{H}_{2}}O}}=\dfrac{{{m}_{{{H}_{2}}O}}}{MolarMas{{s}_{{{H}_{2}}O}}}=\dfrac{33.71g}{18g/mol}=1.87moles$ and
${{n}_{Ca{{C}_{2}}}}=\dfrac{{{m}_{Ca{{C}_{2}}}}}{MolarMas{{s}_{Ca{{C}_{2}}}}}=\dfrac{43.25g}{64g/mol}=0.68moles$

Note:
While finding the limiting reagent, be careful with the number of moles. The formula of no. of moles and molar mass should be remembered. Always start numerical with balancing the chemical equation. If you will not balance the equation the value of mass of the product will be wrong.