Question

$CaC{O_3}\xrightarrow{{heat}}CaO + C{O_2}$
From the above equation calculate the amount of carbon dioxide in grams liberated by heating $25\,g$ of calcium carbonate.

Answer 1

Hint: In a chemical reaction, when one mole of reactant produces one mole of product that means first reactant atomic mass will produce atomic mass of product. So we should know the exact stoichiometric coefficient of chemical reaction of every reactant and product.

Complete step by step solution:
We need a stoichiometric reaction that represents the given change
Here the calcium carbonate decomposes to give calcium oxide and carbon dioxide.
The reaction of decomposition of calcium carbonate can be represented as –
$CaC{O_3}(s)\xrightarrow{{heat}}CaO(s) + C{O_2}(g) \uparrow $
Calcium carbonate calcium oxide carbon dioxide
Here in this case the stoichiometric coefficients are 1 only. We can say 1 mole of calcium carbonate gives 1 mole of carbon dioxide.
But given mass is $25\,g$ of calcium carbonate,
Now, number of moles $ = \dfrac{{given \,mass}}{{molar\, mass}}$
Molar mass of calcium carbonate ( $CaC{O_3}$ ) = molar mass of calcium + molar mass of carbon + 3(molar mass of oxygen
Molar mass of calcium carbonate = 40 +12 + 3(16) = $100\,gmo{l^{ - 1}}$
For $25\,g$of calcium carbonate, number of moles $ = \dfrac{{25\,g}}{{100\,gmo{l^{ - 1}}}} = 0.25\,mol$
According to the stoichiometry of reaction, $0.25\,mol$ of calcium carbonate will give $0.25\,mol$ of carbon dioxide.
Molar mass of carbon dioxide = molar mass of carbon +2(molar mass of oxygen)
Molar mass of carbon dioxide = 12 + 2(16) = $44\,gmo{l^{ - 1}}$
Now $1\,mole$ of carbon dioxide contains $44\,g$ of carbon dioxide.
Therefore, $0.25\,mol$ of carbon dioxide $ = 0.25\,mol \times 44\,gmo{l^{ - 1}} = 11\,g$ of carbon dioxide.

Hence, $25\,g$ of calcium carbonate will produce $11\,g$ of carbon dioxide on decomposition.

Note:
Similarly we can calculate the amount of calcium oxide produced. Since $0.25\,mol$ of calcium carbonate is getting decomposed then $0.25\,mol$ of calcium oxide will produce. And $0.25\,mol$ of calcium oxide contains $ = 56\, gmo{l^{ - 1}} \times 0.025\,mol = 14\,g$ of calcium.