Question

How many grams of $ C{O_2} $ can be prepared from $ 150 $ grams of calcium carbonate $ (CaC{O_3}) $ reacting with an excess of hydrochloric acid $ (HCl) $ solution?
(A) $ 11 $
(B) $ 22 $
(C) $ 33 $
(D) $ 44 $
(E) $ 66 $

Answer 1

Hint :first of all, we need to find out the reaction in which calcium carbonate reacts with hydrochloric acid to form calcium chloride and carbon dioxide and water. Then we will be needed to balance yne reaction and after that comparing their stoichiometric coefficient and then in the Same ratio their moles would be formed. Multiplying moles with their molecular weight so as to get the weight of carbon dioxide formed.

Complete Step By Step Answer:
In the above given question, we are asked the grams of carbon dioxide that will be formed when calcium carbonate reacts with an excess of hydrochloric acid.
Now first let’s write the reaction for the reaction of calcium carbonate and hydrochloric acid: -
 $ CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O $
Now from the above given reaction we can say that one mole of calcium carbonate reacts with two moles of hydrochloric acid to form one mole of calcium chloride, one made of carbon dioxide and one mole of water.
It is given that we have $ 150 $ grams of calcium carbonate, therefore the moles of calcium carbonate we have is equal to the given mass divided by the molecular weight.
 $ moles = \dfrac{{given\,mass}}{{molecular\,weight}} $
 $ \Rightarrow moles = \dfrac{{150}}{{100}} $
 $ \Rightarrow moles = 1.5\,moles $
Now it is given we have $ 1.5 $ moles of calcium carbonate and we have excess of hydrochloric acid which indicates that whole of calcium carbonate is reacted therefore by watching the equation we can say that as we have $ 1.5 $ moles of calcium carbonate, therefore $ 1.5 $ moles of carbon dioxide will be formed.
Now we have $ 1.5 $ moles of carbon dioxide, now to get the mass of carbon dioxide formed we need to multiply the moles of carbon dioxide formed by its molecular weight.
The molecular weight of carbon dioxide is $ 12 + 16 \times 2 = 44 $ grams
Now the mass of carbon dioxide generated is
 $ given\,mass = moles \times molecular\,weight $
 $ \Rightarrow given\,mass = 1.5 \times 44 $
 $ \Rightarrow given\,mass = 66\,grams $
The amount of carbon dioxide formed when $ 150 $ grams of calcium carbonate $ (CaC{O_3}) $ reacting with an excess of hydrochloric acid $ (HCl) $ solution is $ 66 $ grams.

Note :
Calcium carbonate is not very soluble in water. The above given reaction is an acid base reaction where a base calcium carbonate reacts with an acid hydrochloric acid. The carbon dioxide generated in the reaction comes out of the reaction with a different smell.