Question

100 g of $CaC{{O}_{3}}$ is treated with 500 ml M/2 solution of HCl, Find out the volume of $C{{O}_{2}}$ evolved at STP. Which substance is limiting agent
\[CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}\]

Answer 1

Hint: The chemical which is going to be completely consumed when the reaction is completed is called limiting reagent in that reaction. Means those chemicals are taken in exact proportion that is why they won’t be left in a chemical reaction.

Complete step by step answer:
- In the question it is given that calcium carbonate is going to react with hydrochloric acid and asked to find the limiting reagent.
- The given chemical reaction is as follows.
\[CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}\]
- In the above chemical reaction one mole of calcium carbonate reacts with two moles of hydrochloric acid and forms one mole of calcium chloride, one mole of water and one mole of carbon dioxide as the products.
- Mass of calcium carbonate = 100 g
- Molar mass of the calcium carbonate = 100g/mol
- Volume of hydrochloric acid given in the question = 0.5 lit.
- Number of moles of Calcium carbonate = $\dfrac{100}{100}=1mol$
- By using the law of equivalence we can calculate the number of moles of hydrochloric acid.
Number of moles of hydrochloric acid = (Molarity) (volume) = (0.5M) (0.5l) = 0.25 mol.
- From the given chemical reaction 2 moles of HCl = 1 mole of Calcium carbonate
- 0.25 moles of HCl = 0.125 moles of carbon dioxide
- We know that 1 mole at STP = 22.7 liter of volume
 - Therefore 0.125 mole at STP = (22.7) (0.125) = 2.84 liter
- Volume occupied by the carbon dioxide gas is 2.84 liter at STP.

- Therefore the limiting agent in the given chemical reaction is Hydrochloric acid.

Note: Carbon dioxide produced in the given chemical reaction is a gas and occupies 2.84 liter of the volume at STP. By observing all we can say that the hydrochloric acid is consumed totally in the reaction and it is a limiting agent.