To solve this problem, we need to follow these steps:
1. Convert the mass of [tex]\( P_2O_3 \)[/tex] to moles:
The given mass of [tex]\( P_2O_3 \)[/tex] is [tex]\( 92.3 \)[/tex] grams.
The molar mass of [tex]\( P_2O_3 \)[/tex] is [tex]\( 109.94 \)[/tex] grams per mole.
- [tex]\( \text{Moles of } P_2O_3 = \frac{\text{Mass of } P_2O_3}{\text{Molar mass of } P_2O_3} \)[/tex]
- [tex]\( \text{Moles of } P_2O_3 = \frac{92.3 \, \text{g}}{109.94 \, \text{g/mol}} \approx 0.8395 \, \text{mol} \)[/tex]
2. Use stoichiometry to convert moles of [tex]\( P_2O_3 \)[/tex] to moles of [tex]\( H_3PO_3 \)[/tex] based on the balanced chemical equation:
According to the balanced equation:
[tex]\( P_2O_3 + 3H_2O \rightarrow 2H_3PO_3 \)[/tex]
- [tex]\( 1 \text{ mole of } P_2O_3 \text{ produces } 2 \text{ moles of } H_3PO_3 \)[/tex]
- Therefore, [tex]\( \text{Moles of } H_3PO_3 = 0.8395 \, \text{mol} \times 2 \approx 1.6791 \, \text{mol} \)[/tex]
3. Convert the moles of [tex]\( H_3PO_3 \)[/tex] to grams:
The molar mass of [tex]\( H_3PO_3 \)[/tex] is [tex]\( 82.00 \)[/tex] grams per mole.
- [tex]\( \text{Mass of } H_3PO_3 = \text{Moles of } H_3PO_3 \times \text{Molar mass of } H_3PO_3 \)[/tex]
- [tex]\( \text{Mass of } H_3PO_3 = 1.6791 \, \text{mol} \times 82.00 \, \text{g/mol} \approx 137.686 \, \text{g} \)[/tex]
So, filling in the correct numbers in the table, we have:
[tex]\[
\begin{tabular}{l|l|l|l}
$92.3 g P _2 O _3$ & $1 \, \text{mol} \, P _2 O _3$ & $2 \, \text{mol} \, H _3 PO _3$ & $137.686 \, \text{g} \, H _3 PO _3$ \\
\hline
& $109.94 \, \text{g} \, P _2 O _3$ & $1 \, \text{mol} \, P _2 O _3$ & \\
\end{tabular}
\][/tex]
So, the mass of [tex]\( H_3PO_3 \)[/tex] produced is approximately [tex]\( 137.686 \)[/tex] grams.
