Question

How many molecules of $CO_2$ are formed when one milligram of $100\%$ pure $CaCO_3$ is treated with excess hydrochloric acid?

• A. \(6.023 \times 10^{23}\)
• B. \(6.023 \times 10^{21}\)
• C. \(6.023 \times 10^{20}\)
• D. \(6.023 \times 10^{18}\)

Answer 1

The Correct Option is D

$\underset{100 g}{CaCO_{3}}+2HCl \to CaCl_2+H_2O+\underset{\text{1 mol}}{CO_{2}}$
$=6.022 \times 10^{23}$ molecules

$\because \,100 \,g \, CaCO _{3}$ gives, molecules of $C O _{2}$
$=6.1022 \times 10^{23}$
$\therefore \,1 \times 10^{-3} \,g \,CaCO _{3}$ gives molecules of $CO _{2}$
$=\frac{6.022 \times 10^{23} \times 1 \times 10^{-3}}{100} $
$=6.022 \times 10^{18}$

Answer 2

\(CaCO_{3}+2HCl\rightarrow CaCl_{2}+H_{2}O+CO_{2}\)

By seeing the above equation we can say that 1 mole of calcium carbonate is treated with excess hydrochloric acid to form 1 mole of carbon dioxide. We also know that the mass remains conserved in a chemical reaction.

We can calculate Molar mass of \(CaCO_{3}\) as:

\(CaCO_{3}\) : (\(\frac{1\times40g}{mol Ca}\))+(\(\frac{1\times12g}{mol C}\))+(\(\frac{3\times16g}{mol O}\))=100 g/mol \(CaCO_{3}\)

So, now converting the given mass of \(CaCO_{3}\) to moles by dividing by its molar mass.

Moles of \(CaCO_{3}\)= \(\frac{1\times10^{-3}}{100}\)= 1×10^-5mole \(CaCO_{3}\)

So now, we will multiply \(CaCO_{3}\)by the mole ratio between \(CO_{2}\)and \(CaCO_{3}\)from the balanced equation. This will provide us the mol of \(CO_{2}\) in the reaction

\(1\times10^{-3}g\) \(CaCO_{3}\)×\(\frac{1molCaCO_{3}}{100gCaCO_{3}}\)×\(\frac{1molCO_{2}}{1molgCaCO_{3}}\)= 1×10^-5mol \(CO_{2}\)

We know that a mole of any substance contains Avogadro’s number of molecules or atoms.

Hence, applying the unitary method 1×10^-5mol \(CO_{2}\) will give us:

Number of molecules of \(CO_{2}\)= 6.023×1023×1×10^-5mol \(CO_{2}\)

=6.023×1018

Therefore, the correct option is ‘D’.