Excess lead(II) nitrate (Pb(NO3)2) reacts with sodium iodide (NaI) to produce a yellow precipitate.?

Question

1) Excess lead(II) nitrate (Pb(NO3)2) reacts with sodium iodide (NaI) to produce a yellow precipitate. Both solids are dissolved in water before the reaction.
-Write a balanced equation.

2) Explain how to calculate the Average Atomic Mass for a sample if you are given the percent abundance and the isotope mass.

Answer 1

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

2) In order to calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance, expressed as a decimal, and then add the results together

For example using carbon:

Mass #...Weight...%Abundance
12.......12.000000.......98.90
13.......13.003355......1.10

(12.000000) (0.9890) + (13.003355) (0.0110) = 12.011

Hope this helps!

Answer 2

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

Ionically this is

Pb2+(aq) + 2I- (aq) ----> PbI2 (s)

An example should explain the method:
Consider chlrine which has two isotopes of masses 35 and 37 respectively with abundances of 75 and 25 % respectively

Average Atomic mass = Mass due to Cl 35 + Mass due to Cl 37

= 35 x 75 /100 + 25 x 37/100
= 26.25 + 9.25
= 35.50

So in effect you multiply the percentage isotopic abundance by the mass and add the results which give u the ATM.

Answer 3

You probably have a list of solubility rules in your text book that includes "All nitrates are soluble". All alkali metals are soluble, too. So neither of those elements will be components of your precipitate. If your Pb compound has 2 NO3 ions, the charge on Pb is +2. That means it will react with two I- ions. Your balanced equation, then, is
Pb(NO3)2 + 2 NaI -> PbI2(s) + 2Na+ + 2NO3-

Average atomic mass is calculated by converting abundance percentages to decimal numbers (i.e. 96.545% = .96545) , multiplying by the corresponding masses, and adding up your numbers.

Answer 4

I can help you with part 1.

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

How I got this. The valence numbers for lead is +2, nitrate (NO3) is -1, Na is +1, and I is -1.
This is double replacement so the the two metals trade places to get the products shown.
Criss cross the valence numbers to get the correct subscripts and then add the coefficients to balance and you will get teh answer shown.

Answer 5

Write the equation which shows that AgNO3 produces AgCl i a 1:1 molar ratio. So convert 75g of AgNO3 into moles (amount = mass/molar mass) and, due to the 1:1 ratio this will equal the number of moles of AgCl that will be produced so rearrange the expression and use the molar mass of AgC to convert this amount into a mass.

Answer 6

blood and other flammable liquids