Answer:
Step-by-step explanation:
Moles of Pb(NO3)2 = mass/molecular mass
= 50.0 grams/(207.20*1 + 14.01*2 + 16*6)
= 50.0 grams/331.22
= 0.15 moles
Moles of NaI
= 30/(22.99+126.9)
= 30/149.89
= 0.2 Moles
A. NaI is less 2x Pb(NO3)2 so NaI is the limiting reagent.
B. The ratio is 1 to 2 so there is 0.15 - 0.2/2 = 0.05 mole
or 16.78 grams of Pb(NO3)2 left.
C. As NaI is limiting, only 0.2 Moles of NaNO3 is formed.
Mass = Moles * Molecular Mass
Molecular Mass of NaNO3 can be calculated as:
Na - 22.99
N - 14.01
O - 3(16) = 48
23+14+48 = 85gram / mole
Thus, Mass = 0.2*85 = 17 gram of NaNO3
Mass is conserved in a chemical reaction.
Mass of PbI2 can be calculated as:
50+30-16.78-17
= 46.3 gram of PbI2
Mass =
12.75
Thus, 12.75g of Sodium Nitrate can be formed