Question: If 20 g of \[CaC{O_3}\] is treated with 20 g of \[HCl\] , how many grams of can be generated \[C{O_2...

Question

If 20 g of $CaC{O_3}$ is treated with 20 g of $HCl$ , how many grams of can be generated $C{O_2}$ according to the following equation?
$CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}$
(Molecular masses: $CaC{O_3}$ =100u, $HCl$ =36.5u, $C{O_2}$ =44u)

Answer 1

Solution

Hint: We can easily find the total amount of $C{O_2}$ gas generated by just finding out the limiting reagent in the reaction $CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}$ .

Complete step by step solution:
In order to calculate the total amount of $C{O_2}$ gas generated in the balanced chemical reaction $CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}$ following steps have to be followed:
Step 1: First, we will convert the given amounts into moles.
Mass of $CaC{O_3}$ given = 20 g
Moles of $CaC{O_3}$ $= \dfrac{{{\text{mass of }}CaC\mathop O\nolimits_3 }}{{{\text{molecular mass of}}CaC\mathop O\nolimits_3 }} = \dfrac{{20}}{{100}} = 0.2$ .
Mass of $HCl$ given = 20 g
Moles of $HCl$$$$ = \dfrac{{20}}{{36.5}} = 0.547$ .
Step 2: In this step we have to identify limiting reagent.
Limiting reagent is defined as those substances which get completely consumed in a chemical reaction. They also govern the quantity of the final product of a chemical reaction.
From the above balanced equation, we can say that 1 mol of $CaC{O_3}$ reacts with 2 mol of $HCl$ .
Therefore, 0.2 mol of $CaC{O_3}$ will react with 0.4 mol of $HCl$ .
As, we have 0.547 of $HCl$ , hence $CaC{O_3}$ is the limiting reagent and $HCl$ is the excessive reagent
Step 3: In this last step, we will calculate the amount of $C{O_2}$ formed
As the amount of product formed depends on limiting reagent, hence, we can calculate amount of $C{O_2}$ formed as follows : 1 mol of $CaC{O_3}$ form $C{O_2}$ = 1mol
Therefore, 0.2 mol of $CaC{O_3}$ will form $C{O_2}$ = $1 \times 0.2 = 0.2$$$$mol$ .
Since, molecular mass of $C{O_2}$ is 44 u
Therefore, the mass of $C{O_2}$ formed $= {\text{ }}0.2 \times 44 = $$$$8.8{\text{ }}g$ .

Note: It should be remembered that the chemical reaction should be balanced while finding limiting reagent. Also, the amount of product form in a chemical reaction is limited by limiting reagent.