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Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

Starting with with 50.0 grams of Pb(NO3)2 and 30.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of the excess reactant remains?
C. How many grams of each product is formed?
D. If 12 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?

User Manikawnth
by
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2 Answers

3 votes

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

User Nmvictor
by
3.0k points
3 votes

Answer:

Step-by-step 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%

User Fankibiber
by
3.4k points