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 $$={\displaystyle \frac{(663.1202\times {10}^{23})}{(602.8233\times {10}^{23})}}\times 2.5$$
                     $$=\left(1.1\right)\times 2.5$$
                    = 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 $$6.022\times {10}^{23}$$ .