White phosphorus reacts with hot aqueous NaOH: P4+3NaOH+3H2OrightarrowPH3+3NaH2PO2. In an experiment, 12.4 g of P4 (molar mass 124 g mol^-1) reacts…

Moles of $\mathrm{P_4}=12.4/124=0.100$ mol. From the balanced equation (check: P 4 = 1 + 3, Na 3, O 6 = 6, H 9 = 3 + 6), each mole of $\mathrm{P_4}$ gives 3 mol of $\mathrm{NaH_2PO_2}$ and 1 mol of $\mathrm{PH_3}$.

Mass of $\mathrm{NaH_2PO_2}=0.100\times3\times88=26.4$ g.

Oxidation states: in $\mathrm{P_4}$ each P is $0$; in $\mathrm{NaH_2PO_2}$, P is $+1$ (oxidation, 3 atoms per $\mathrm{P_4}$, losing 3 electrons in total); in $\mathrm{PH_3}$, P is $-3$ (reduction, 1 atom, gaining 3 electrons). Electron balance: $3\times1=3$ lost $=1\times3=3$ gained. Electrons gained $=0.100\times3=0.30$ mol.

Option 1 uses only one mole of product; option 2 multiplies the electron count by 3 (counting all P atoms); option 3 takes 88 g as the mass for the whole $\mathrm{P_4}$ run, ignoring the amounts of substance. The answer is $26.4$ g and $0.3$ mol.