Question

A test tube along with calcium carbonate in it initially weighed \[{\text{30}}{\text{.08 g}}\] . A heating experiment was performed on this test tube till calcium carbonate completely decomposed with evolution of a gas. Loss of weight during this experiment was \[{\text{4}}{\text{.40 g}}\]. What is the weight of the empty test tube in this experiment?
(A) \[{\text{20}}{\text{.08g}}\]
(B) \[{\text{21}}{\text{.00 g}}\]
(C) \[{\text{24}}{\text{.50 g}}\]
(D) \[{\text{2}}{\text{.008 g}}\]

Answer 1

Hint: Use the reaction stoichiometry to convert the loss of weight to the initial mass of calcium carbonate. From the initial mass of the test tube and calcium carbonate, subtract the mass of calcium carbonate to obtain the mass of the empty test tube.

Complete step by step answer:
Let \[x{\text{ g}}\] be the weight of the empty test tube.
Thermal decomposition of calcium carbonate gives calcium oxide and carbon dioxide. Write the balanced chemical reaction for the thermal decomposition of calcium carbonate.
\[{\text{CaC}}{{\text{O}}_3}{\text{ }} \to {\text{ CaO + C}}{{\text{O}}_2}\]
The molecular weights of calcium carbonate and carbon dioxide are \[{\text{100 g/mol}}\] and \[{\text{44 g/mol}}\] respectively. One mole of calcium carbonate, on thermal decomposition gives one mole of carbon dioxide. Hence, \[{\text{100 g}}\] of calcium carbonate, on thermal decomposition will give \[{\text{44 g}}\] of carbon dioxide gas.
Calculate the amount of calcium carbonate needed to produce \[{\text{4}}{\text{.4 g}}\] of carbon dioxide gas.
\[\dfrac{{100}}{{44}} \times 4.4 = 10{\text{ g}}\]
Thus, the mass of calcium carbonate before thermal decomposition is \[10{\text{ g}}\] .
Calculate the mass of the empty test tube
\[{\text{x = 30}}{\text{.08}} - {\text{10}}{\text{.00 = 20}}{\text{.08 g}}\]
Hence, the mass of the empty test tube is \[{\text{20}}{\text{.08 g}}\] .

Hence, the correct option is the option (A).

Note: A decomposition reaction is one in which one molecule breaks into two (or more) smaller molecules. In the above reaction, one molecule of calcium carbonate breaks into two small molecules of calcium oxide and carbon dioxide. This decomposition is carried out in presence of heat. Hence, we call it thermal decomposition reaction.