Question

What mass in miligram of $CaCO_{3}$ is required to react completely with 25 ml of 0.75 M HCl according to the reaction $CaCO_{3}\left(\right.s\left.\right)+2HCl\left(\right.a\left.\right) \rightarrow CaCl_{2}\left(\right.a\left.\right)+CO_{2}\left(\right.g\left.\right)+H_{2}O\left(\right.l\left.\right)$ ?

Answer 1

Given

$\mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Moles present in 25 ml of 0.75 M HCl $=\frac{0.75}{1000}\times 25$

$=0.01875$ moles

Since 2 moles react completely with $\text{1} \, \text{mole} \, \text{CaCO}_{\text{3}} \, \text{=} \, \text{100} \, \text{g}$

0.01875 moles react completely with $\text{CaCO}_{\text{3}} = \frac{100}{2} \times 0.01875$

$0.9375 \, \text{g} = 937.5 \, \text{mg}$