Q4.137P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Two aqueous solutions contain the ions indicated below.

 

(a) Write balanced molecular, total ionic, and net ionic equations for the reaction that occurs when the solutions are mixed. (b) If each sphere represents 0.050 mol of ion, what mass (in g) of precipitate forms, assuming 100% reaction? (c) What is the concentration of each ion in solution after reaction?

Step-by-Step Solution

Answer

Answer: 

a. The balanced molecular equation for the reaction is, 

$N{a}_{2}C{O}_3\left(aq\right)+CaC{l}_2\left(aq\right)\to CaC{O}_3\left(s\right)+2NaCl\left(aq\right)$

The total ionic equation for the above reaction is,

$2N{a}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)+C{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)\to CaC{O}_3\left(s\right)+2N{a}^{+}\left(aq\right)+2C{l}^{-}\left(aq\right)$

 The net ionic equation is,

 $C{O}_3^{2-}\left(aq\right)+C{a}^{2+}\left(aq\right)\to CaC{O}_3\left(s\right)$

b. The mass of the precipitate (CaCO₃) is 10.0 g.

 c. Thus, the concentration of Na⁺, Cl^-, CO₃^2-, and Ca²⁺ are 0.60 M, 0.40 M, 0.10 M and O M, respectively.

 

Step 1: (a) Determination of equations

$$\begin{gathered} \text{The balanced molecular equation for the reaction is, } \\ N{a}_{2}C{O}_3\left(aq\right)+CaC{l}_2\left(aq\right)\to CaC{O}_3\left(s\right)+2NaCl\left(aq\right) \\ \text{ The total ionic equation for the above reaction is, } \\ 2N{a}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)+C{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)\to CaC{O}_3\left(s\right)+2N{a}^{+}\left(aq\right)+2C{l}^{-}\left(aq\right) \\ \text{ The net ionic equation is, } \\ C{O}_3^{2-}\left(aq\right)+C{a}^{2+}\left(aq\right)\to CaC{O}_3\left(s\right) \end{gathered}$$

 

 

 

Step 2: (b) Determination of mass of precipitate

The number of red spheres (CO₃^2-) in the first beaker are 3.

The number of brown spheres (Na⁺) in the first beaker are 6.

The number of green spheres (Cl^-) in the second beaker are 4.

The number of blue spheres (Ca²⁺) in the second beaker are 2.

 

Each sphere is 0.050 moles.

 

Number of moles of CO₃^2- are,

$$\begin{gathered} =3\times 0.050mol\backslash \mathrm{hfill} \\ =0.15mol \end{gathered}$$ 

Number of moles of Na⁺ are,

 $$\begin{gathered} =6\times 0.050mol \\ =0.30mol \end{gathered}$$

 

Number of moles of Cl^- are,

 $$\begin{gathered} =4\times 0.050mol \\ =0.20mol \end{gathered}$$

 

Number of moles of Ca²⁺ are,

 $$\begin{gathered} =2\times 0.050mol \\ =0.10mol \end{gathered}$$

 

For the formation of precipitate 1 mol of Ca²⁺ reacts with 1 mol of CO₃^2-.

 $C{O}_3^{2-}\left(aq\right)+C{a}^{2+}\left(aq\right)\to CaC{O}_3\left(s\right)$

 

The limiting reagent in this reaction is Ca²⁺. 

1 mol of Ca²⁺ produces 1 mol of CaCO₃. So, 0.10 mol of Ca²⁺ will produces 0.10 mol of CaCO₃.

 

Now, the mass of CaCO₃ is,

 $$\begin{gathered} Mass=moles\times molar mass \\ =0.10mol\times 100.09g/mol \\ =10.0g \end{gathered}$$

 

Thus, the mass of the precipitate (CaCO₃) is 10.0 g.

 

Step 3: (c) Determination of concentration of each ion

Total volume of the solution is,

 $$\begin{gathered} =250.0mL+250.0mL \\ =500.0mL \\ =0.500L \end{gathered}$$

 

As Na⁺ and Cl^- ions does not undergo any reaction, there will be no change in the moles.

 

Concentration of Na⁺ is,

 $$\begin{gathered} Concentration=\frac{moles}{volume} \\ =\frac{0.30mol}{0.500L} \\ =0.60M \end{gathered}$$

 

Concentration of Cl^- is,

 $$\begin{gathered} Concentration=\frac{moles}{volume} \\ =\frac{0.20mol}{0.500L} \\ =0.40M \end{gathered}$$

 

 

Initial moles of CO₃^2- is 0.15 mol. 0.10 mol of this ion will form precipitate. SO, the moles left are 0.05 mol.

Concentration of CO₃^2- are,

 $$\begin{gathered} Concentration=\frac{moles}{volume} \\ =\frac{0.05mol}{0.500L} \\ =0.10M \end{gathered}$$

 

Ca²⁺ is the limiting reagent, it will be completely consumed in the reaction. So, the concentration of Ca²⁺ is 0.

 

Thus, the concentration of Na⁺, Cl^-, CO₃^2-, and Ca²⁺ are 0.60 M, 0.40 M, 0.10 M and O M, respectively.