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: Analyze phosphorus atoms

Let the coefficient of $P_4$ be $1$. In $PH_3$ and $Na_2HPO_3$, phosphorus has different oxidation - states. The total number of phosphorus atoms in products should equal the number of phosphorus atoms in the reactant $P_4$.

Step2: Set up equations for other elements

Let the coefficient of $NaOH$ be $a$, the coefficient of $H_2O$ be $b$, the coefficient of $PH_3$ be $x$, and the coefficient of $Na_2HPO_3$ be $y$.
For phosphorus: $4=x + y$ (since there are 4 phosphorus atoms in $P_4$).
For sodium: $a = 2y$.
For hydrogen: $a+2b = 3x + y$.
For oxygen: $a + b=3y$.
By trial - and - error or solving the system of equations, when $x = 3$ and $y = 1$.
If $y = 1$, then from $a = 2y$, $a = 2$.
From $a + b=3y$, substituting $a = 2$ and $y = 1$ gives $2 + b=3$, so $b = 1$.

Step3: Write the balanced equation

$P_4(s)+3NaOH(aq)+3H_2O(l)
ightarrow 3PH_3(g)+3Na_2HPO_3(aq)$

Answer:

$P_4(s)+3NaOH(aq)+3H_2O(l)
ightarrow 3PH_3(g)+3Na_2HPO_3(aq)$