Chapter F, Problem H.10E

(a)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction given using the shortcut.

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{H}}_{\text{2}}\text{O}(l)$

Concept Introduction:

Chemical equation is a short form representation of a chemical reaction with all the required conditions necessary for the reaction to take place.  As the atoms are neither created nor destroyed in a chemical reaction, the number of atoms of same element has to be equal on both sides of the equation.  This is known as balanced chemical equation.  Coefficients can be changed but not the subscript.

Answer to Problem H.10E

Balanced chemical equation is ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The given chemical equation is shown below;

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing phosphate ion:  In the reactant side, there is one phosphate ion while on the product side, there are two phosphate ions.  Adding coefficient $2$ before ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ balances the phosphate ions on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing calcium atom:  In the reactant side, there is one calcium atom while on the product side, there are three calcium atoms.  Adding coefficient $3$ before ${\text{Ca(OH)}}_{\text{2}}$ balances the calcium atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing hydrogen atoms:  In the reactant side, there are twelve hydrogen atoms while on the product side, there are only two hydrogen atoms.  Adding coefficient $6$ before ${\text{H}}_{\text{2}}\text{O}$ balances the hydrogen atoms on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)$

(b)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction given using the shortcut.

    ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+{\text{HClO}}_{\text{2}}(aq)\to {\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$

Concept Introduction:

Refer part (a).

Answer to Problem H.10E

Balanced chemical equation is ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+6{\text{HClO}}_{\text{2}}(aq)\to 2{\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$.

Explanation of Solution

The given chemical equation is shown below;

    ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+{\text{HClO}}_{\text{2}}(aq)\to {\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$

Balancing sulfate ion:  In the reactant side, there are three sulfate ions while on the product side, there is only one sulfate ion.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ balances the sulfate ions on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+{\text{HClO}}_{\text{2}}(aq)\to {\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$

Balancing hydrogen atom:  In the reactant side, there is one hydrogen atom while on the product side, there are six hydrogen atoms.  Adding coefficient $6$ before ${\text{HClO}}_{\text{2}}$ balances the hydrogen atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+6{\text{HClO}}_{\text{2}}(aq)\to {\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$

Balancing chlorite ion:  In the reactant side, there are six chlorite ions while on the product side, there are only three chlorite ions.  Adding coefficient $2$ before ${\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}$ balances the chlorite ions on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{Cr}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+6{\text{HClO}}_{\text{2}}(aq)\to 2{\text{Cr(ClO}}_{\text{2}}{\text{)}}_{\text{3}}(aq)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)$