Question

$Ca{Cl}_2\left(aq\right)+{Na}_2{CO}_3\left(aq\right)\to Ca{CO}_3\left(s\right)+2NaCl\left(aq\right)$
$I$. $2.50g$ of $Ca{Cl}_2$ is fully dissolved in a beaker of water and $2.50g$ of ${Na}_2{CO}_3$ is fully dissolved in water in a second beaker. The two solutions are mixed to form a $Ca{CO}_3$ precipitate and aqueous $NaCl.{Na}_2{CO}_3$ will be the limiting reactant in this experiment.
$II$. Sodium carbonate has a molar mass which is less than the molar mass of calcium chloride.

• A. Statement $I$ is true, Statement $II$ is true
• B. Statement $I$ is true, Statement $II$ is false
• C. Statement $I$ is false, Statement $II$ is true
• D. Statement $I$ is false, Statement $II$ is false

Answer 1

The correct option is C Statement $I$ is false, Statement $II$ is true

$\text{Given:}\left(CaC{l}_2\right(aq)+N{a}_{2}C{O}_3(aq)⟶CaC{O}_3(s)+2\mathrm{NaCl}(aq)$

$\begin{array}{c}\text{Initial moles of}{CaCl}_2=\frac{2\cdot 50}{111}\text{mol.}\\ \text{ands Initial moles of}{Na}_2{CO}_3=\frac{2.50}{106}mol\text{.}\\ \text{Hence,}{CaCl}_2\text{is acting as limiting reagent.}\\ \text{Molar mass of sodium carbonate is less than that of calcium chloride.}\end{array}$

Option C is correct answer