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Question: When the calcium carbonate is reacted with an excess amount of hydrochloric acid, how much \(CaCO_3\...

When the calcium carbonate is reacted with an excess amount of hydrochloric acid, how much CaCO3CaCO_3 is required to produce 11.2 litres of CO2CO_2 at STP?
(Given molar masses: CaCO3CaCO_3 = 100g/mol, HCl= 36.5g/mol, CO2CO_2 = 44.0g/mol)
A) 25.0g
B) 44.0g
C) 50.0g
D) 100.0g
E) None of the above

Explanation

Solution

Volume of one mole of any gas is called molar volume and is equal to 22.4 L at STP. Molar volume allows conversion between moles and volume of gas at STP.

Complete answer:
We know
CaCO3(aq) + 2HCl(aq)CaCl2(aq) + H2O (g) + CO2(g){\text{CaC}}{{\text{O}}_3}{\text{(aq) + 2HCl(aq)}} \to {\text{CaC}}{{\text{l}}_{\text{2}}}{\text{(aq) + }}{{\text{H}}_{\text{2}}}{\text{O (g) + C}}{{\text{O}}_2}(g)
The above chemical equation is stoichiometrically balanced
At S.T.P
22.4L of CO2CO_2 = 1 mol of CO2CO_2
So 11.2L of CO2CO_2 = 0.5 mol of CO2CO_2
1 mol of CO2CO_2 requires 1 mol of CaCO3CaCO_3
So, 0.5 mol of CO2CO_2 will require 0.5 mol of CaCO3CaCO_3
According to mole concept

1 mol of CaCO3 = molar mass of CaCO3 0.5 mol of CaCO3 = 100 × 0.50.5 mol of CaCO3 = 50g{{\text{1 mol of CaC}}{{\text{O}}_3}{\text{ = molar mass of CaC}}{{\text{O}}_3}} \\\ {{\text{0}}{\text{.5 mol of CaC}}{{\text{O}}_{\text{3}}}{\text{ = 100 }} \times {\text{ 0}}{\text{.5}}} {{\text{0}}{\text{.5 mol of CaC}}{{\text{O}}_{\text{3}}}{\text{ = 50}}g}

Hence, the answer to the above question is option C.

Note: S.T.P - Standard Temperature and Pressure are standard arrangements of conditions for experimental estimations to be set up to permit correlations to be made between different sets of data.
Since 1982, STP has been defined as a temperature of 273.15 K and an absolute pressure of exactly 10510^5 Pa.
Mole - In the SI system, mole (symbol, mol) was introduced as seventh base quantity for the amount of a substance. 1 mol contains as many particles/entities as there are atoms in exactly 12 g of the 12C isotope. No. of entities in 1 mol = 6.02×10236.02 \times10 ^{-23} .
It is called Avogadro constant (NAN_A).

Molar mass - The mass of 1 mol of a substance in grams is called its molar mass. The molar mass in grams is numerically equal to atomic/molecular mass in u.
1 mol = gram formula mass (molar mass)
= 6.02×10236.02 \times10 ^{-23} particles
= 22.4L of a gas at STP .
The molar mass of a polymer is directly related to its properties. As the molecular weight increases, mechanical properties generally increase. Every polymer has an ideal molecular weight at which the balance of different properties (such as processability, strength, brittleness etc.) is optimized.