Question

56.75 g of $HCl$ is dissolved in 2 L of water. Calculate the mass of $CaC{O}_3$ required to react completely with 25 mL of this $HCl$ solution.
$CaC{O}_3+2HCl\to CaC{l}_2+C{O}_2+H_{2}O$

• A. 10.54 g
• B. 9.275 g
• C. 0.9715 g
• D. 0.09727 g

Answer 1

The correct option is C 0.9715 g

We know,
$Molarityofsolution=\frac{molesofsolute}{volumeofsolutioninlitre}$
$Molarity=\frac{\frac{56.75}{36.5}}{2}=0.777M$
Again we know,
$Numberofmoles=volume\times molarity$
Therefore, number of moles of $HCl$ present in 25 mL of solution of $HCl$
$=\frac{25}{1000}\times 0.777=19.43\times {10}^{-3}mol$
From the balanced chemical reaction,
$CaC{O}_3+2HCl\to CaC{l}_2+C{O}_2+H_{2}O$
2 mole $HCl$ react with 1 mole of $CaC{O}_3$.
$\therefore 19.43\times {10}^{-3}$ moles of HCl will react with = $\frac{19.43}{2}\times {10}^{-3}molofCaC{O}_3$
Therefore the required weight of $CaC{O}_3$
$=\frac{19.43}{2}\times {10}^{-3}mol\times 100g/mol=0.9715g$