Question

Pb(NO3)2 + 2KCI -> PbCl2 + 2K(NO3)

10 g of Pb(NO3)2 and 6 g of KCl

Find the theoretical yield PbCl2 and K(NO3)

Answer 1

The only thing I need help on is the fact that the products aren't 1 mole each.

Answer 2

How many moles of the reactant do we have and what is the limiting reagent?

Answer 3

.03 moles of Pb(NO3)2 and .08 moles of KCI. Limiting is Pb(NO3)2 because .06 KCl is needed but we have .08 of it.

Answer 4

You're doing great so far!

So now you know your limiting reagent. This is what gets used up.

Now all we have to do is use the moles to mole ratio, which is 1:2:1:2

This means that if we use 1 mole of Pb(NO3)2, we would need 2 moles of KCl. And together, it would form 1 mole of PbCl2 and 2 moles of K(NO3).

Now, just replace that 1 with 0.03 moles and keep the ratios the same.

0.03 moles of Pb(NO3)2 will react with __ moles of KCl to form __ moles of PbCl2 and __ moles of K(NO3)

Answer 5

0.03 moles of Pb(NO3)2 will react with 2 moles of KCl to form 1 moles of PbCl2 and 2 moles of K(NO3)

Answer 6

no no no

what you just wrote is a ratio of 0.03 to 2 to 1 to 2

Answer 7

if we have 0.03 moles, how many moles of KCl will react with it?

The ratio is 1 to 2

Answer 8

ohh .06

Answer 9

yes, and so what about PbCl2 and K(NO3)

Answer 10

0.03 moles of Pb(NO3)2 will react with .06 moles of KCl to form .03 moles of PbCl2 and .06 moles of K(NO3)

Answer 11

ta-da!

Answer 12

now that wasn't that hard, was it? :))

Answer 13

Lesss gooo and no it wasn't xd

Answer 14

I just realized that I have been overthinking all these types of problems and thanks for helping me with like 1000 questions today

Answer 15

Great work!!

Answer 16

It was my pleasure! :D

Answer 17

Thanks and great help

Answer 18

Just don't forget to use the molar mass to convert back to grams!

Answer 19

Yup thanks for the reminder

Answer 20

Of course!