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Question: \(CaC{O_3}\xrightarrow{{heat}}CaO + C{O_2}\) From the above equation calculate the amount of carbo...

CaCO3→heatCaO+CO2CaC{O_3}\xrightarrow{{heat}}CaO + C{O_2}
From the above equation calculate the amount of carbon dioxide in grams liberated by heating 25 g25\,g of calcium carbonate.

Explanation

Solution

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 –
CaCO3(s)→heatCaO(s)+CO2(g)↑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 g25\,g of calcium carbonate,
Now, number of moles =given massmolar mass = \dfrac{{given \,mass}}{{molar\, mass}}
Molar mass of calcium carbonate ( CaCO3CaC{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 gmol−1100\,gmo{l^{ - 1}}
For 25 g25\,gof calcium carbonate, number of moles =25 g100 gmol−1=0.25 mol = \dfrac{{25\,g}}{{100\,gmo{l^{ - 1}}}} = 0.25\,mol
According to the stoichiometry of reaction, 0.25 mol0.25\,mol of calcium carbonate will give 0.25 mol0.25\,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 gmol−144\,gmo{l^{ - 1}}
Now 1 mole1\,mole of carbon dioxide contains 44 g44\,g of carbon dioxide.
Therefore, 0.25 mol0.25\,mol of carbon dioxide =0.25 mol×44 gmol−1=11 g = 0.25\,mol \times 44\,gmo{l^{ - 1}} = 11\,g of carbon dioxide.

**Hence, 25 g25\,g of calcium carbonate will produce 11 g11\,g of carbon dioxide on decomposition.

Note: **
Similarly we can calculate the amount of calcium oxide produced. Since 0.25 mol0.25\,mol of calcium carbonate is getting decomposed then 0.25 mol0.25\,mol of calcium oxide will produce. And 0.25 mol0.25\,mol of calcium oxide contains =56 gmol−1×0.025 mol=14 g = 56\, gmo{l^{ - 1}} \times 0.025\,mol = 14\,g of calcium.