Question

Macmillan Learning

For the chemical reaction
2 KI + Pb(NO3)2 →→→ Pbl₂ + 2 KNO3
what mass of lead(II) iodide is produced from 4.91 mol of potassium iodide?
mass:

Answer 1

1132 g

Step-by-step explanation:

since you're trying to find the amount (grams?) you get from one element/compound from another, you need to do a stoichiometry equation.

1. Usually, you need to go from grams, then you need to find the moles of that substance. After you get the mole amount, you need to do a mole : mole ratio, that way you "switch" the elements/compounds you're using. Once you get that amount, it's it moles. Therefore you need to convert those moles into a gram amount (in this case using lead(II) iodide)

First, you start with your 4.91 mol of KI - you don't need to go from gram to mole since you're already given your mol amount.

Then, you do a mole : mole ratio -

• 4.91 mol KI ×
  `(1 mol PbI2)/(2 mol KI) \\` --

1. What you try to cancel goes on the bottom, so if you cancel both the KI's, you're left with the element you're looking for -- in this case lead (II) iodide.
2. How you find the amount of moles it possesses, and how you know which number you use for a mole : mole ratio is by the giant numbers in front.

• KI = 2 moles
• Pb(NO3)2 = 1 mole
• PbI2 = 1 mole
• KNO3 = 2 mole

Once you do your mole : mole ratio, you end in moles. If the question is looking for moles, you're done. But, if you're looking for grams you need to do one more step -

1. Find the molar mass of your compound you're looking for (PbI2)

• Pb = 207
• I2 = ~127(x2, since there is two Iodine's) = 252 g --> 207+254= ~461g

Then you start doing your equation -

`4.91 mol * (1 mol PbI2)/(2 mol KI) * (461 g PbI2)/(1 mol PbI2)`

1. you multiply this by 461/1mol because for every mole of PbI2, there is 461 grams.
2. You put the 1 mol on bottom because you want to get rid of moles, so you put the unit you want to get rid of on the bottom. Leaving just grams as the end.