Answer
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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.
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.
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