Calculate the number of molecules of carbon (IV) oxide produced when 10g of CaCO3 is treated with 100cm of
0.2mol dm HCl. The equation of the reaction is
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(1) + CO2(g)
ICa=40 (=12 O=161

The mass of produced is 0.44 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

......(1)

Given mass of calcium carbonate = 10 g

Molar mass of calcium carbonate = 100 g/mol

Plugging values in equation 1:

The formula used to calculate molarity:

.....(2)

Molarity of HCl =     (Conversion factor:

Volume of solution =          (Conversion factor:

Putting values in equation 2, we get:

For the given chemical reaction:

By stoichiometry of the reaction:

If 2 moles of HCl reacts with 1 mole of calcium carbonate

So, 0.02 moles of HCl will react with = of calcium carbonate

As the given amount of calcium carbonate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, HCl is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of produces 1 mole of

So, 0.02 moles of will produce = of

We know, molar mass of = 44 g/mol

Putting values in equation 1, we get:

Hence, the mass of produced is 0.44 g

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