Answer:
mass PbI₂ formed = 1383 grams
Explanation:
Pb(NO₃)₂ + 2NaI => 2NaNO₃ + PbI₂(s)
6 mol NaI => 1/2(6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)
1) Calculate the number of mols,n, of the substance
n = mass/ molar mass = 326.0 g / 58.45 g / mol = 5.577 moles
2) Calculate the molar heat of fusion as the total heat released by the sample divided by the number of moles
hf = heat released / n = 4325.8 cal / 5.577 moles = 775. 59 cal /mol
The valence electrons should be the correct answer
Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,

Where, m is mass of iron
s is specific heat of iron
d T is change in temperature in celcius

If
1 cal = 4.2 J
Then,

Thus 0.389 k cal of enrgy is required by a 15.6 g Fe to reach to 355 C^0