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

Last updated date: 14th Sep 2024
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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 $6.022 \times {10^{23}}$ .