Chapter SH, Problem 19.2APP

To determine

To balance: The given equation.

Answer to Problem 19.2APP

The balanced equation is:

  $3\text{Zn}{\left(\text{OH}\right)}_2+2{\text{H}}_3{\text{PO}}_4\to {\text{Zn}}_3{\left({\text{PO}}_4\right)}_2+6{\text{H}}_2\text{O}$

Explanation of Solution

Given:

The following equation:

  $\underset{\_}{}\text{Zn}{\left(\text{OH}\right)}_2+\underset{\_}{}{\text{H}}_3{\text{PO}}_4\to \underset{\_}{}{\text{Zn}}_3{\left({\text{PO}}_4\right)}_2+\underset{\_}{}{\text{H}}_2\text{O}$

Calculation:

Count and list the atoms found on both sides of the equation:

  $\text{Zn}{\left(\text{OH}\right)}_2+{\text{H}}_3{\text{PO}}_4\to {\text{Zn}}_3{\left({\text{PO}}_4\right)}_2+{\text{H}}_2\text{O}$

  

The Zinc atoms are unbalanced. Put a 3 in front of Zn. The hydrogen atoms are now $6+3=9$ on the right hand side and 2 on the left hand side. Put a 2 in front of ${\text{H}}_3$ and 6 in front of ${\text{H}}_2$ . The hydrogen atoms are now 12 on both sides. This also balances oxygen atoms and the phosphorus atoms. The balanced equation looks like:

  $3\text{Zn}{\left(\text{OH}\right)}_2+2{\text{H}}_3{\text{PO}}_4\to {\text{Zn}}_3{\left({\text{PO}}_4\right)}_2+6{\text{H}}_2\text{O}$

Conclusion:

The balanced equation is:

  $3\text{Zn}{\left(\text{OH}\right)}_2+2{\text{H}}_3{\text{PO}}_4\to {\text{Zn}}_3{\left({\text{PO}}_4\right)}_2+6{\text{H}}_2\text{O}$