Question

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

Answer 1

Answer: The mass of
`CO_2` produced is 0.44 g

Step-by-step explanation:

• For calcium carbonate:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

Given mass of calcium carbonate = 10 g

Molar mass of calcium carbonate = 100 g/mol

Plugging values in equation 1:

`\text{Moles of calcium carbonate}=(10g)/(100g/mol)=0.1 mol`

• For HCl:

The formula used to calculate molarity:

`\text{Molarity of solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (mL)}}` .....(2)

Molarity of HCl =
`0.2mol/dm^3=0.2mol/L` (Conversion factor:
`1L=1dm^3`

Volume of solution =
`100cm^3=100mL` (Conversion factor:
`1mL=1cm^3`

Putting values in equation 2, we get:

`0.2=\frac{\text{Moles of HCl}* 1000}{100}\\\\\text{Moles of HCl}=(0.2* 100)/(1000)=0.02mol`

For the given chemical reaction:

`CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)`

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 =
`(1)/(2)* 0.02=0.01mol` 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
`HCl` produces 1 mole of
`CO_2`

So, 0.02 moles of
`HCl` will produce =
`(1)/(2)* 0.02=0.01mol` of
`CO_2`

We know, molar mass of
`CO_2` = 44 g/mol

Putting values in equation 1, we get:

`\text{Mass of }CO_2=(0.01mol* 44g/mol)=0.44g`

Hence, the mass of
`CO_2` produced is 0.44 g