Chapter 3, Problem 3.16PAE

3.16 Balance the following equations.

(a) reaction to produce "super-phosphate" fertilizer:

Ca₃(PO₄)₂(s)+H₂SO₄(aq)( Ca(H₂PO₄)₂(aq)+CaSO₄(s)

(b) reaction to produce diborane, B₂H₆:

NaBH₄(s) + H₂SO₄(aq) ( B₂H₆(g) + H₂(g) + Na₂SO₄(aq)

(c) reaction to produce tungsten metal from tungsten(VI)

oxide:

WO₃(s)+ H2(g) ( W(s) + H₂O(l)

(d) decomposition of ammonium dichromate:

(NH₄)₂Cr₂O₇(s)(N₂(g)+H₂O(l)+Cr₂O₃(s)

Interpretation Introduction

To balance:

reaction to produce "superphosphate" fertilizer:

The equation:

$\begin{array}{l}\\ {\text{Ca}}_3{{\text{(PO}}_{\text{4}}}_{\text{)}}{\text{(s) + H}}_2{\text{SO}}_4(\text{aq})\to {\text{ Ca(H}}_2{\text{PO}}_4{\text{)}}_2{\text{ (aq) + CaSO}}_4\text{ (s)}\end{array}$

Explanation of Solution

We need to find the stoichiometric coefficient of reactants and products in order to the sum of the mass of the reactants is equal to the sum of the mass of the products. First, we need to balance the atoms that are not hydrogen or oxygen. After that, we have to balance the hydrogens and finally the oxygens. So that, the balanced chemical equation is:

${\text{Ca}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2{\text{(s) + 2H}}_2{\text{SO}}_4(\text{aq})\to {\text{ Ca(H}}_2{\text{PO}}_4{\text{)}}_2{\text{ (aq) + 2CaSO}}_4\text{ (s)}$

Interpretation Introduction

To balance: The equation:

reaction to produce diborane,

$\begin{array}{l}{\text{ B}}_2{\text{H}}_6\text{: }\\ {\text{NaBH}}_4{\text{(s)+ H}}_2{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{B}}_2{\text{H}}_6{\text{ (g) + H}}_2{\text{(g)+Na}}_2{\text{SO}}_4\text{ (aq)}\end{array}$

Explanation of Solution

We need to find the stoichiometric coefficient of reactants and products in order to the sum of the mass of the reactants is equal to the sum of the mass of the products. First, we need to balance the atoms that are not hydrogen or oxygen. After that, we have to balance the hydrogens and finally the oxygens. So that, the balanced chemical equation is:

${\text{2NaBH}}_4{\text{(s)+ H}}_2{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{B}}_2{\text{H}}_6{\text{ (g) + 2H}}_2{\text{(g)+Na}}_2{\text{SO}}_4\text{ (aq)}$

Interpretation Introduction

To balance: The equation:

reaction to produce tungsten metal from tungsten(VI) oxide: 

$\begin{array}{l}\\ {\text{WO}}_{\text{3}}{\text{(s) + H}}_2(g)\to {\text{ W (s)+ H}}_{\text{2}}\text{O (l)}\end{array}$

Explanation of Solution

We need to find the stoichiometric coefficient of reactants and products in order to the sum of the mass of the reactants is equal to the sum of the mass of the products. First, we need to balance the atoms that are not hydrogen or oxygen. After that, we have to balance the hydrogens and finally the oxygens. So that, the balanced chemical equation is:

${\text{2WO}}_{\text{3}}{\text{(s) + 6 H}}_2(g)\to {\text{ 2W (s)+ 6H}}_{\text{2}}\text{O (l)}$

Interpretation Introduction

To balance: The equation:

decomposition of ammonium dichromate:

$\begin{array}{l}\\ {\text{ (NH}}_4{\text{)}}_2{\text{Cr}}_2{\text{O}}_7\text{(s)}\to {\text{ Cr}}_2{\text{O}}_3{\text{ (s) + N}}_2{\text{(g)+ H}}_2\text{O (l)}\end{array}$

Explanation of Solution

We need to find the stoichiometric coefficient of reactants and products in order to the sum of the mass of the reactants is equal to the sum of the mass of the products. First, we need to balance the atoms that are not hydrogen or oxygen. After that, we have to balance the hydrogens and finally the oxygens. So that, the balanced chemical equation is:

${\text{2(NH}}_4{\text{)}}_2{\text{Cr}}_2{\text{O}}_7\text{(s)}\to {\text{ 2Cr}}_2{\text{O}}_3{\text{ (s) + 2N}}_2{\text{(g)+ 8H}}_2\text{O (l)}$

Conclusion

The balanced equations are:

$\begin{array}{l}{\text{a) Ca}}_3{{\text{(PO}}_{\text{4}}}_{\text{)}}{\text{(s) + 2H}}_2{\text{SO}}_4(\text{aq})\to {\text{ Ca(H}}_2{\text{PO}}_4{\text{)}}_2{\text{ (aq) + 2CaSO}}_4\text{ (s)}\end{array}$

$\begin{array}{l}{\text{b) 2NaBH}}_4{\text{(s)+ H}}_2{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{B}}_2{\text{H}}_6{\text{ (g) + 2H}}_2{\text{(g)+Na}}_2{\text{SO}}_4\text{ (aq)}\end{array}$

$\begin{array}{l}{\text{c) 2WO}}_{\text{3}}{\text{(s) + 6 H}}_2(g)\to {\text{ 2W (s)+ 6H}}_{\text{2}}\text{O (l)}\end{array}$

$\begin{array}{l}{\text{d) 2(NH}}_4{\text{)}}_2{\text{Cr}}_2{\text{O}}_7\text{(s)}\to {\text{ 2Cr}}_2{\text{O}}_3{\text{ (s) + 2N}}_2{\text{(g)+ 8H}}_2\text{O (l)}\end{array}$