Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. p₄(s) + naoh(aq) + h₂o(l) → ph₃(g) + na₂hpo₃(aq)

Explanation:

Step1: Balance phosphorus atoms

Let the coefficient of $P_4$ be $1$. In $PH_3$ and $Na_2HPO_3$, the sum of phosphorus - atom contributions from the products must equal 4 (since there are 4 P - atoms in $P_4$). Let the coefficient of $PH_3$ be $x$ and the coefficient of $Na_2HPO_3$ be $y$, then $x + y=4$.

Step2: Consider oxidation - reduction and charge balance

The oxidation state of P in $P_4$ is 0, in $PH_3$ is - 3, and in $Na_2HPO_3$ is + 3. By the principle of conservation of charge (or oxidation - reduction), we can set up equations based on the change in oxidation states. After some trial - and - error and considering the balance of other elements (Na, H, O), we find that $x = 3$ and $y = 1$.

Step3: Balance sodium atoms

Since the coefficient of $Na_2HPO_3$ is 1, the coefficient of $NaOH$ should be 3 to balance the sodium atoms.

Step4: Balance hydrogen and oxygen atoms

With the coefficients of $P_4 = 1$, $NaOH=3$, $PH_3 = 3$, $Na_2HPO_3 = 1$, the coefficient of $H_2O$ is 3 to balance the hydrogen and oxygen atoms.
The balanced chemical equation is $P_4(s)+3NaOH(aq)+3H_2O(l)
ightarrow3PH_3(g)+Na_2HPO_3(aq)$

Answer:

$P_4(s)+3NaOH(aq)+3H_2O(l)
ightarrow3PH_3(g)+Na_2HPO_3(aq)$