Question

Q87. $4.5$moles of calcium carbonate are reacted with dilute hydrochloric acid

1. Write the equation for the reaction
2. What is the mass of $4.5$ moles of calcium carbonate(Relative molecular mass of calcium carbonate is $100$.
3. What is the volume of carbon dioxide liberated at STP.
4. What mass of calcium chloride is formed (Relative molecular mass of calcium chloride is $111$
5. How many moles of $\mathrm{HCl}$ are used in this reaction..

Answer 1

(i)

• The balanced reaction when calcium carbonate reacts with dil. hydrochloric acid :
  $\underset{\mathrm{calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{calcium}\mathrm{chloride}}{{\mathrm{CaCl}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)\uparrow }$

(ii)

• Calculate the mass of calcium carbonate :
  we know $1$ mole of calcium carbonate weighs $100g$
  Similarly, $4.5$moles of calcium carbonate weighs $$\begin{gathered} 100\times 4.5 \\ =450g \end{gathered}$$
• The resulting mass of calcium carbonate is $450g$.

(iii)

• Calculate the volume of carbon dioxide liberated :
  We know $1$ mole of calcium carbonate liberates $22.4l$
  Similarly, $4.5$moles of calcium carbonate liberates $$\begin{gathered} 22.4\times 4.5 \\ =100.8L \end{gathered}$$
• The resulting volume of calcium carbonate is $100.8L$

(iv)

• Calculate the mass of calcium chloride formed :
  We know from the balanced equation $100g$of calcium carbonate forms$111g$ of calcium chloride
  Similarly, $450g$of calcium carbonate will formed from, $$\begin{gathered} \frac{111}{100}\times 450 \\ =499.5gofcalciumchloride. \end{gathered}$$
• Hence the resulting mixture contains $499.5g$ of calcium chloride.

(v)

• Calculate the moles of Hydrochloric acid used in the reaction :
  For $1$mole of calcium carbonate $2$moles of hydrochloric acid is used
  Similarly, $4.5$moles of calcium carbonate uses $$\begin{gathered} 4.5\times 2 \\ =9molesofHCl \end{gathered}$$
• The moles of hydrochloric acid required is $9$ moles.