Answer:
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
Answer:
Explanation:
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
MM for each compound -
Pb(NO3): 207 + 14x2 + 16x3x2 = 331
PI2: 207 + 127x2 = 461
NaI: 23 + 127 = 150
NaNO3: 23 + 14 + 16x3 = 85
Moles of Pb(NO3)2 = 50/331 = 0.15
Moles of NaI = 30/150 = 0.2
Ratio of moles is 1:2
So NaI is limiting
Limited to 0.2/2 = 0.1 mole of Pb(NO3)2
Excess = 0.15 - 0.1 = 0.05 mole
Mass remains = 0.05x331 = 16.55 grams
Moles of NaNO3 formed = Moles of NaI reacted = 0.2
Mass = 0.2x85 = 17 grams
Moles of PbI2 formed = Moles of Pb(NO3)2 reacted = 0.1
Mass = 0.1x461 = 46.1 grams
If 12 grams of NaNO3 actually formed in the reaction,
percent yield = 12/17x100% = 70.6%
