Question

balance the chemical equation below using the smallest possible whole number stoichiometri

p4(s) + naoh(aq) + h2o(l) → ph3(g) + na2hpo3(aq)

Explanation:

Step1: Assign oxidation states

• In \(P_4\), the oxidation state of \(P\) is \(0\).
• In \(PH_3\), the oxidation state of \(P\) is \(- 3\).
• In \(Na_2HPO_3\), the oxidation state of \(P\) is \(+ 3\).

Step2: Determine the change in oxidation states

• For \(P\) going from \(0\) (in \(P_4\)) to \(-3\) (in \(PH_3\)): change is \(-3\).
• For \(P\) going from \(0\) (in \(P_4\)) to \(+3\) (in \(Na_2HPO_3\)): change is \(+3\).

Step3: Balance the oxidation - reduction part

• Let the coefficient of \(PH_3\) be \(x\) and of \(Na_2HPO_3\) be \(y\).
• From the oxidation - state changes, since the total increase in oxidation state must equal the total decrease.
• For \(P_4\) (4 \(P\) atoms), if we consider the ratio of \(PH_3\) and \(Na_2HPO_3\) based on oxidation - state changes.
• Let's assume \(x = 1\) and \(y = 1\) for the \(P\) atoms in \(PH_3\) and \(Na_2HPO_3\). But since we have \(P_4\) (4 \(P\) atoms), we find that for the redox part: \(P_4

ightarrow PH_3+Na_2HPO_3\).

• The ratio of \(PH_3\) to \(Na_2HPO_3\) based on electron transfer (since the change in oxidation state per \(P\) atom for \(PH_3\) is \(-3\) and for \(Na_2HPO_3\) is \(+3\)) and considering the \(P_4\) (4 \(P\) atoms), we get \(P_4

ightarrow 2PH_3 + 2Na_2HPO_3\) (because \(2\times(- 3)+2\times(+3)=0\) for the net change in oxidation state for the \(4\) \(P\) atoms from \(P_4\)).

Step4: Balance \(Na\) atoms

• Since we have \(2Na_2HPO_3\), we need \(4NaOH\) to balance the \(Na\) atoms.

Step5: Balance \(H\) and \(O\) atoms

• The equation becomes \(P_4+4NaOH + 2H_2O=2PH_3+2Na_2HPO_3\)

Answer:

\(P_{4}(s)+4NaOH(aq)+2H_{2}O(l)=2PH_{3}(g)+2Na_{2}HPO_{3}(aq)\)