Question

 $CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}$
The mass of calcium chloride formed when 2.5 g of Calcium Carbonate is dissolved in excess of Hydrochloric acid is:

Answer 1

Hint: According to the equation, one mole of \[CaC{O_3}\] produces one mole of \[CaC{l_2}\].Molar mass is the sum of the atomic masses of elements present in the compound.The number of molecules in one gram-mole of a substance, defined as the molecular weight in grams, is \[6.023 \times {10^{23}}\], a quantity called the Avogadro constant.

Complete step by step solution:
The molecular mass of \[CaC{O_3}\] is = 100.0869
The molecular mass of \[CaC{l_2}\] is = 110.098
1 Mole of any substance is equals to \[6.023 \times {10^{23}}\] particles

In the above reaction, if we take 1gram of \[CaC{O_3}\] it will contain \[602.8233 \times {10^{23}}\] molecules and similarly 1 mole of \[CaC{O_3}\] will produce 1 mole of CaCl2 that is \[663.1202 \times {10^{23}}\] molecules while reacts with hydrochloride.

Since we use 2.5 g of CaCO3 \[ = \dfrac{{(663.1202 \times {{10}^{23}})}}{{(602.8233 \times {{10}^{23}})}} \times 2.5\]
                     \[ = \left( {1.1} \right) \times 2.5\]
                    = 2.75 g
The 2.75 g mass of calcium chloride formed while reacting 2.5 g Calcium Carbonate with excess hydrochloride.

Note: One mole is defined as the amount of substance containing the same number of entities as the number of atoms in a sample of pure \[^{12}C\] weighing exactly 12 grams. The number of entities composing a mole has been experimentally determined to be \[6.022 \times {10^{23}}\] .