Question

How would you balance ${\mathrm{PCl}}_5+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{HCl}$

Answer 1

Balancing the Chemical equation:

• When we balance chemical equations, we want the same number of each sort of atom on both sides of the equation.
• Only the coefficients are changed. The numbers in front of the molecule are called coefficients.
• Subscripts should never be changed.
• Metals must be balanced first, followed by nonmetals such as oxygen and hydrogen.
• The unbalanced chemical reaction provided is

$\underset{\mathrm{Phosphrous}\mathrm{Pentachloride}}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphoric}\mathrm{acid}}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{Chloride}}{\mathrm{HCl}\left(\mathrm{l}\right)}$

• It is a double displacement redox process.
• The following table shows the number of atoms on each side:

Reactant side	Product side
1 Phosphorus atom from ${\mathrm{PCl}}_5$	1 Phosphorus atom from ${\mathrm{H}}_3{\mathrm{PO}}_4$
2 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$	4 Hydrogen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$ and $\mathrm{HCl}$
5 Chlorine atoms from ${\mathrm{PCl}}_5$	1 Chlorine atom from $\mathrm{HCl}$
1 Oxygen atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	4 Oxygen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$

• As there are five Chlorine atoms on the left side, we must put a $5$ before $\mathrm{HCl}$ on the right side.

$\underset{\mathrm{Phosphrous}\mathrm{Pentachloride}}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphoric}\mathrm{acid}}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{Chloride}}{5\mathrm{HCl}\left(\mathrm{l}\right)}$

Reactant side	Product side
1 Phosphorus atom from ${\mathrm{PCl}}_5$	1 Phosphorus atom from ${\mathrm{H}}_3{\mathrm{PO}}_4$
2 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$	8 Hydrogen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$ and $\mathrm{HCl}$
5 Chlorine atoms from ${\mathrm{PCl}}_5$	5 Chlorine atom from $\mathrm{HCl}$
1 Oxygen atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	4 Oxygen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$

• Second, Hydrogen must be balanced. As shown on the left, there is a total of eight Hydrogen atoms; thus, a $4$ must be inserted before Water on the right side to balance them.

$\underset{\mathrm{Phosphrous}\mathrm{Pentachloride}}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}+4\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphoric}\mathrm{acid}}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{Chloride}}{5\mathrm{HCl}\left(\mathrm{l}\right)}$

• Finally, the oxygens must be balanced, but now, in this case, they are already at $4$ on each side; therefore, the equation is now balanced.

Reactant side	Product side
1 Phosphorus atom from ${\mathrm{PCl}}_5$	1 Phosphorus atom from ${\mathrm{H}}_3{\mathrm{PO}}_4$
8 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$	8 Hydrogen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$ and $\mathrm{HCl}$
5 Chlorine atoms from ${\mathrm{PCl}}_5$	5 Chlorine atom from $\mathrm{HCl}$
4 Oxygen atom from ${\mathrm{H}}_2{\mathrm{SO}}_4$	4 Oxygen atoms from ${\mathrm{H}}_3{\mathrm{PO}}_4$

• Thus, the balanced chemical reaction is:

$\underset{\mathrm{Phosphrous}\mathrm{Pentachloride}}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphoric}\mathrm{acid}}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrogen}\mathrm{Chloride}}{5\mathrm{HCl}\left(\mathrm{l}\right)}$