Question

By heating ${\text{10g}}$of ${\text{CaC}}{{\text{O}}_{\text{3}}}$,${\text{5}}{\text{.6g}}$
of ${\text{CaO}}$ is formed. What is the weight of ${\text{C}}{{\text{O}}_{\text{2}}}$ obtained in this reaction?

A.${\text{2}}{\text{.2g}}$
B.${\text{2}}{\text{.3g}}$
C.${\text{3}}{\text{.2g}}$
D. ${\text{4}}{\text{.4g}}$

Answer 1

Hint:: To determine the number of moles and grams of any reactant or product balanced equation is required. After writing the balanced equation, by comparing the gram amount of reactant and product, the amount of carbon dioxide can be determined. We can determine the number of gram weights by using the mole formula.

Formula used: ${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete step by step solution:
The breaking of any compound in two or more smaller compounds on heating is known as decomposition.
Decomposition, of calcium carbonate in calcium oxide and carbon dioxide is as follows:
\[{\text{CaC}}{{\text{O}}_3}{\text{(s)}} \to {\text{CaO(s)}} + \,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,\,\]
According to the balanced reaction, the decomposition of one moles of calcium carbonate, gives one mole of calcium oxide, and one mole of carbon dioxide.
We will use the mole formula to determine the gram of calcium carbonate, carbon dioxide in one mole as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of the calcium carbonate is $100\,{\text{g/mol}}$.
Substitute $100\,{\text{g/mol}}$ for molar mass and $1$ mol for moles of calcium carbonate.
${\text{1}}\,{\text{mol}}\,\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{100\,{\text{g/mol}}}}$
${\text{Mass}}\,{\text{of CaC}}{{\text{O}}_3}\,{\text{ = }}\,{\text{1}}\,{\text{mol}}\,\, \times 100\,{\text{g/mol}}$
${\text{Mass}}\,{\text{of CaC}}{{\text{O}}_3}\,{\text{ = }}\,100\,{\text{g}}$
So, one mole of carbon dioxide is equal to $100$ gram.

Molar mass of the carbon dioxide is $44\,{\text{g/mol}}$.
Substitute $44\,{\text{g/mol}}$ for molar mass and $1$ mol for moles of carbon dioxide.
${\text{1}}\,{\text{mol}}\,\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{44\,{\text{g/mol}}}}$
${\text{Mass}}\,{\text{of C}}{{\text{O}}_2}\,{\text{ = }}\,{\text{1}}\,{\text{mol}}\,\, \times 44\,{\text{g/mol}}$
${\text{Mass}}\,{\text{of C}}{{\text{O}}_2}\,{\text{ = }}\,44\,{\text{g}}$
So, one mole of carbon dioxide is equal to $44$ gram.
So, according to the balanced equation, stoichiometry of calcium carbonate and carbon dioxide is one so, $100$ gram calcium carbonate will give $44$gram carbon dioxide.
Compare the gram ratio to determine the gram amount of carbon dioxide formed by ${\text{10g}}$ calcium carbonate as follows:
${\text{100}}\,{\text{g}}\,{\text{CaC}}{{\text{O}}_{\text{3}}}\, = \,{\text{44}}\,{\text{g}}\,{\text{C}}{{\text{O}}_2}$
${\text{10}}\,{\text{g}}\,{\text{CaC}}{{\text{O}}_{\text{3}}}\, = \,{\text{4}}{\text{.4}}\,{\text{g}}\,{\text{C}}{{\text{O}}_2}$
So, $10$ gram of calcium carbonate ${\text{CaC}}{{\text{O}}_{\text{3}}}$ will give ${\text{4}}{\text{.4}}$ gram carbon dioxide${\text{C}}{{\text{O}}_2}$.
Therefore, option (D) ${\text{4}}{\text{.4g}}$ is correct.

Note: Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary. Molar mass of any compound is determined by adding the atomic mass of each constituting atom. The atomic mass is the sum of numbers of protons and neutrons.