The mass of \( \text{PbI}_2 \) formed from 0.77 moles of \( \text{Pb(NO}_3)_2 \) reacting with \( \text{KI} \) is approximately 360.96 grams.
To find the mass of \( \text{PbI}_2 \) formed, we first use the balanced chemical equation to establish the mole ratio between \( \text{Pb(NO}_3)_2 \) and \( \text{PbI}_2 \).
The balanced equation is:
\[ \text{Pb(NO}_3)_2 (\text{aq}) + 2 \text{KI} (\text{aq}) \rightarrow \text{PbI}_2 (\text{s}) + 2 \text{KNO}_3 (\text{aq}) \]
The mole ratio is 1:1 between \( \text{Pb(NO}_3)_2 \) and \( \text{PbI}_2 \).
Given moles of \( \text{Pb(NO}_3)_2 \): 0.77 moles
Therefore, moles of \( \text{PbI}_2 \) formed: 0.77 moles
Now, use the molar mass of \( \text{PbI}_2 \) (molar mass of Pb + 2 * molar mass of I):
\[ \text{Molar mass of PbI}_2 = 207.2 \, \text{g/mol} + 2 \times 126.9 \, \text{g/mol} \]
Calculate the mass:
\[ \text{Mass of PbI}_2 = 207.2 + 2 \times 126.9 \, \text{g/mol} \times 0.77 \, \text{moles} \]
\[ \text{Mass of PbI}_2 \approx 360.96 \, \text{g} \]