91Ó°ÊÓ

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

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

The total ionic equation of the balanced molecular equation is like

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

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

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

Answer of subpart (b):

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

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

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

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

Answer of subpart (c):

Answer: You need to identify the oxidizing agent

The balanced chemical equation is

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

Oxidation no of oxygen is -2.

Oxidation no of Mn in MnO₄^- is +7.

Oxidation no of Mn in Mn²⁺ is +2.

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 chemical equation is

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

Oxidation no of oxygen is -2.

Oxidation no of C in H₂C₂O₄ is +3.

Oxidation no of C in CO₂ is +4.

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

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,

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

Now, moles of MnO₄^-

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

Hence, moles of MnO₄^- are 0.003839mol.

According to the balanced equation

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

2 moles of MnO₄^- reacts with 5 moles H₂C₂O₄

Now, moles of H₂C₂O₄

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

Hence, moles of H₂C₂O₄ are 0.0095975mol.

According to the question

1 mole of CaCl₂ produces 1 mole H₂C₂O₄

Now, moles of CaCl₂

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

Hence, moles of CaCl₂ are 0.0095975mol.

Moles of CaCl₂ = 0.0095975moles

Molecular mass of CaCl₂ = 110g/mol

Again you know,

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

Mass of CaCl₂

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

Hence, mass of CaCl₂ is 1.0557g.

Again you know,

Mass percent of CaCl₂

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

Hence, mass percent of CaCl₂ is 54.56%.

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