Question

Mixtures of CaCl₂ and NaCl are used to melt ice on roads. A dissolved 1.9348-g sample of such a mixture was analyzed by using excess Na₂C₂O₄ to precipitate the Ca²⁺ as CaC₂O₄. The CaC₂O₄ was dissolved in sulfuric acid, and the resulting H₂C₂O₄ was titrated with 37.68mL of 0.1019M KMnO₄ solution.

(a) Write the balanced net ionic equation for the precipitation reaction.

(b) Write the balanced net ionic equation for the titration reaction. (See Sample Problem 4.11.)

(c) What is the oxidizing agent?

(d) What is the reducing agent?

(e) Calculate the mass percent of CaCl₂ in the original sample.

Short answer

Answer of subpart (a):

Answer: You need to write the balanced net ionic equation for the precipitation reaction.

Answer of subpart (b):

Answer: You need to write the balanced net ionic equation for the titration reaction.

Answer of subpart (c):

Answer: You need to identify the oxidizing agent

Answer of subpart (d):

Answer: You need to identify the reducing agent

Answer of subpart (e):

Answer: You need to calculate the mass percent of CaCl₂ in the original sample.

Step-by-step solution

Molecular equation for precipitation reaction

Mixture of Na₂C₂O₄ reacts with CaCl₂ produces CaC₂O₄ and NaCl. The molecular balanced equation is

${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+{\mathrm{CaCl}}_2\left(\mathrm{aq}\right)\to {\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)+2\mathrm{NaCl}\left(\mathrm{aq}\right)$

Total ionic equation for precipitation reaction

The total ionic equation of the balanced molecular equation is like

$2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)+2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Net ionic equation for precipitation reaction

The Na⁺ and Cl^- ions are spectator ions. Hence, they will not be present in the net ionic equation. Therefore, the net ionic equation is like

${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)$

Answer of subpart (b):

Answer: You need to write the balanced net ionic equation for the titration reaction.

Molecular equation for titration reaction

${\mathrm{K}}^{+}\left(\mathrm{aq}\right)+{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{K}}^{+}\left(\mathrm{aq}\right)$

Net ionic equation for titration reaction 

The K⁺ ion is spectator ion. Hence, it will not be present in the net ionic equation. Therefore, the net ionic equation is like

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+10{\mathrm{CO}}_2\left(\mathrm{g}\right)+8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Answer of subpart (c):

Answer: You need to identify the oxidizing agent

The balanced equation

The balanced chemical equation is

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+10{\mathrm{CO}}_2\left(\mathrm{g}\right)+8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Calculation of oxidation nos.

Oxidation no of oxygen is -2.

Oxidation no of Mn in MnO₄^- is +7.

Oxidation no of Mn in Mn²⁺ is +2.

Oxidizing agent

Oxidation no of Mn in MnO₄^- is +7 decreases to +2 in Mn²⁺. Therefore, MnO₄^- undergoes reduction, hence acts as an oxidizing agent.

Answer of subpart (d):

Answer: You need to identify the reducing agent

The balanced equation

The balanced chemical equation is

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+10{\mathrm{CO}}_2\left(\mathrm{g}\right)+8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Calculation of oxidation nos.

Oxidation no of oxygen is -2.

Oxidation no of C in H₂C₂O₄ is +3.

Oxidation no of C in CO₂ is +4.

Reducing agent

Oxidation no of C in H₂C₂O₄ is +3 increases to +4 in CO₂. Therefore, H₂C₂O₄undergoes oxidation, hence acts as a reducing agent.

Calculation of moles of MnO4-

Answer of subpart (e):

Answer: You need to calculate the mass percent of CaCl₂ in the original sample.

According to the question,

Volume of MnO₄^- = 37.68mL = 0.03768L

Molarity of MnO₄^- = 0.1019M

Again you know,

$\mathrm{moles}=\mathrm{molarity}\times \mathrm{volume}$

Now, moles of MnO₄^-

$\begin{array}{l}=(0.1019\times 0.03768)\left(\mathrm{mol}\right)\\ =0.003839\left(\mathrm{mol}\right)\end{array}$

Hence, moles of MnO₄^- are 0.003839mol.

Calculation of moles of H2C2O4

According to the balanced equation

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+10{\mathrm{CO}}_2\left(\mathrm{g}\right)+8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

2 moles of MnO₄^- reacts with 5 moles H₂C₂O₄

Now, moles of H₂C₂O₄

$\begin{array}{l}=0.003839\times \frac{5}{2}\left(\mathrm{mol}\right)\\ =0.0095975\left(\mathrm{mol}\right)\end{array}$

Hence, moles of H₂C₂O₄ are 0.0095975mol.

Calculation of moles of CaCl2

According to the question

1 mole of CaCl₂ produces 1 mole H₂C₂O₄

Now, moles of CaCl₂

^{$\begin{array}{l}=0.0095975\times \frac{1}{1}\left(\mathrm{mol}\right)\\ =0.0095975\left(\mathrm{mol}\right)\end{array}$}

Hence, moles of CaCl₂ are 0.0095975mol.

Calculation of mass of CaCl2

Moles of CaCl₂ = 0.0095975moles

Molecular mass of CaCl₂ = 110g/mol

Again you know,

$\mathrm{mass}=\mathrm{moles}\times \mathrm{mass}\left(\mathrm{molar}\right)$

Mass of CaCl₂

_{$\begin{array}{l}=0.0095975\times 110\left(\mathrm{g}\right)\\ =1.0557\left(\mathrm{g}\right)\end{array}$}

Hence, mass of CaCl₂ is 1.0557g.

Calculation of mass percent of CaCl2

Again you know,

Mass percent of CaCl₂

$\begin{array}{l}=\frac{\mathrm{mass}\left({\mathrm{CaCl}}_2\right)}{\mathrm{mass}\left(\mathrm{sample}\right)}\times 100\mathrm{\%}\\ =\frac{1.0557}{1.9348}\times 100\mathrm{\%}\\ =54.56\mathrm{\%}\end{array}$

Hence, mass percent of CaCl₂ is 54.56%.

Conclusion

Hence, mass percent of CaCl₂ in the original sample is 54.56%.