
200 grams of marble chips are dropped into 100g of hydrochloric acid. How much chips will remain undissolved? What weight of anhydrous calcium chloride and what weight of $CO_2$ gas could be obtained from it?
${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{ }} \to {\text{ CaO + }}{\text{ C}}{{\text{O}}_2}$
Answer
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Hint: We must remember that the limiting agent in this reaction is HCl. First find the moles of the both $CaCO_3$ and HCl after finding that form the chemical equation required for this question and solve accordingly.
Complete step by step answer:
Let’s start with discussing the steps that we will take to solve this question for easier understanding. First we will need to find the moles of the marble chips (calcium carbonate). For finding the moles of calcium carbonate we will divide the mass by the molecular weight of calcium carbonate. Substituting the value of mass of calcium carbonate and molecular mass of calcium carbonate we get,
Moles of calcium carbonate = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass of Calcium Carbonate}}}} = \dfrac{{200g}}{{100}}$= $2$moles.
After this we will find the moles of HCl which will be calculated in similar way,
Moles of HCl = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass of HCl}}}} = \dfrac{{100g}}{{36.5g}}$= $2.74$moles.
After getting the moles of both the compounds we will formulate the equation
${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{ + 2HCl }} \to {\text{ CaC}}{{\text{l}}_2}{\text{ + }}{\text{ C}}{{\text{O}}_2}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
This gives the following data:
0.63 moles of chips will remain undissolved. Which means 63g of $CaC{O_3}$will be undissolved.
Also $1.37$ moles of $CaC{l_2}$ and $C{O_2}$ is being formed which means 152.67 g $152.67g$ $CaC{l_2}$ and $60.28g$ of $C{O_2}$ is being formed.
Note:
We must remember that the production of $C{O_2}$ can be confirmed by doing the lime test by passing the gas released through the lime water. If the lime water turns milk white in colour then $C{O_2}$ is released in the reaction. Also the limiting agent is the compound which limits the reaction by getting totally depleted itself.
Complete step by step answer:
Let’s start with discussing the steps that we will take to solve this question for easier understanding. First we will need to find the moles of the marble chips (calcium carbonate). For finding the moles of calcium carbonate we will divide the mass by the molecular weight of calcium carbonate. Substituting the value of mass of calcium carbonate and molecular mass of calcium carbonate we get,
Moles of calcium carbonate = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass of Calcium Carbonate}}}} = \dfrac{{200g}}{{100}}$= $2$moles.
After this we will find the moles of HCl which will be calculated in similar way,
Moles of HCl = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass of HCl}}}} = \dfrac{{100g}}{{36.5g}}$= $2.74$moles.
After getting the moles of both the compounds we will formulate the equation
${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{ + 2HCl }} \to {\text{ CaC}}{{\text{l}}_2}{\text{ + }}{\text{ C}}{{\text{O}}_2}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
This gives the following data:
0.63 moles of chips will remain undissolved. Which means 63g of $CaC{O_3}$will be undissolved.
Also $1.37$ moles of $CaC{l_2}$ and $C{O_2}$ is being formed which means 152.67 g $152.67g$ $CaC{l_2}$ and $60.28g$ of $C{O_2}$ is being formed.
Note:
We must remember that the production of $C{O_2}$ can be confirmed by doing the lime test by passing the gas released through the lime water. If the lime water turns milk white in colour then $C{O_2}$ is released in the reaction. Also the limiting agent is the compound which limits the reaction by getting totally depleted itself.
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