Some Basic Concepts of Chemistry

Let's first write the balanced chemical equation for the reaction between calcium carbonate ($\mathrm{CaCO}_3$) and hydrochloric acid ($\mathrm{HCl}$).

$ \mathrm{CaCO}_3 + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 $

From the equation, we can see that 1 mole of $\mathrm{CaCO}_3$ reacts with 2 moles of $\mathrm{HCl}$ to produce 1 mole of $\mathrm{CO}_2$.

Now, let's calculate the moles of $\mathrm{CaCO}_3$. The molar mass of $\mathrm{CaCO}_3$ is:

$ \text{Molar mass of} \ \mathrm{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol} $

Given that we have $20 \ \mathrm{g}$ of $\mathrm{CaCO}_3$, the number of moles of $\mathrm{CaCO}_3$ is:

$ \text{Moles of} \ \mathrm{CaCO}_3 = \frac{20 \ \mathrm{g}}{100 \ \mathrm{g/mol}} = 0.2 \ \mathrm{mol} $

Now, let's determine the moles of $\mathrm{HCl}$ in the given solution. The solution is $20\%$ $\mathrm{HCl}$ by weight, and we have $100 \ \mathrm{mL}$ of the solution. Assuming the density of the solution is approximately $1 \ \mathrm{g/mL}$, we get:

$ \text{Mass of} \ \mathrm{HCl} \ \text{solution} = 100 \ \mathrm{mL} \times 1 \ \mathrm{g/mL} = 100 \ \mathrm{g} $

Since the solution is $20\%$ $\mathrm{HCl}$, the mass of $\mathrm{HCl}$ is:

$ \text{Mass of} \ \mathrm{HCl} = 20\% \ \text{of} \ 100 \ \mathrm{g} = 20 \ \mathrm{g} $

The molar mass of $\mathrm{HCl}$ is:

$ \text{Molar mass of} \ \mathrm{HCl} = 1 + 35.5 = 36.5 \ \mathrm{g/mol} $

The number of moles of $\mathrm{HCl}$ is:

$ \text{Moles of} \ \mathrm{HCl} = \frac{20 \ \mathrm{g}}{36.5 \ \mathrm{g/mol}} \approx 0.548 \ \mathrm{mol} $

The reaction requires 2 moles of $\mathrm{HCl}$ for 1 mole of $\mathrm{CaCO}_3$. Since we have only $0.2 \ \mathrm{mol}$ of $\mathrm{CaCO}_3$, we need:

$ \text{Required moles of} \ \mathrm{HCl} = 0.2 \ \mathrm{mol} \times 2 = 0.4 \ \mathrm{mol} $

We have more than enough $\mathrm{HCl}$ ($0.548 \ \mathrm{mol}$ available) to fully react with $0.2 \ \mathrm{mol}$ of $\mathrm{CaCO}_3$.

Therefore, the amount of $\mathrm{CO}_2$ produced will be equal to the moles of $\mathrm{CaCO}_3$ reacted, which is $0.2 \ \mathrm{mol}$.

To find the mass of $\mathrm{CO}_2$ produced, we use its molar mass:

$ \text{Molar mass of} \ \mathrm{CO}_2 = 12 + 2 \times 16 = 44 \ \mathrm{g/mol} $

The mass of $\mathrm{CO}_2$ is:

$ \text{Mass of} \ \mathrm{CO}_2 = 0.2 \ \mathrm{mol} \times 44 \ \mathrm{g/mol} = 8.80 \ \mathrm{g} $

The amount of $\mathrm{CO}_2$ produced is $8.80 \ \mathrm{g}$.

Given the options, the correct answer is:

Option B - 8.80 g

MS₂ = M + 32 $\times$ 2 = M + 64

% of sulphur = $\left(\frac{64}{M+64}\right)\times100=40.06$

M + 64 = $\frac{6400}{40.06}$

M + 64 = 160

M = 160 $-$ 64 = 96 u

Molecular mass

$ = {{Weight\,of\,organic\,substance\,taken} \over {Air\,displaced\,at\,STP}} \times 22400$

$ = {{0.2} \over {56}} \times 22400 = 80$

Molarity = ${{10 \times density\, \times \,percentage\,weight\,of\,solute} \over {molecular\,weight\,of\,the\,solute}}$

$\Rightarrow$ Density $ = {{3.60 \times 98} \over {10 \times 29}} = 1.216 = 1.22$