Chem help Please!!?~~~pressed for time and need to understand!?

Question

How many grams of PbCl2 will be formed from 70 mL of 1.5 M KCl and excess Pb(NO3)2? Given the reaction:Pb(NO3)2+ 2KCL gives PbCl2+ 2KNO3

I think I need to find the molar mass and use that to convert but other than that, I'm pretty clueless where to go. This is an online college class, so Im pretty much on my own. Thanks for help!!

Answer 1

Alright, no worries!!

You need to understand the stoichiometric relationship between Pb(NO3)2 and PbCl2 is a 1 : 1 ratio and KCl and PbCl2 is a 2 : 1 ratio.

Therefore, we can progress from here. This is a rather simple problem if you know how to break it down.

We also know we have an excess of Pb(NO3)2 and therefore this is NOT a limiting reagent problem.

You know you have 70 mL of 1.5 KCl M. Therefore you know by M= mol/ L that      

1.5 M x .070L will give you the moles  i.e. 1.5 x  (?mol/ .070L)

The moles: .105 of KCl

Now you have an edge.

You start with .105 moles KCl x (1 mole PbCl2 / 2 moles KCl ) = .0525 moles PbCl2

This cancels out moles of KCl while leaving you with the amount of moles of PbCl2

You did not have to stop there, but I wanted to make it easier on you if you don't understand. I merely used the stoichiometric relationship.

Now you can also start with this answer to obtain grams. You need the molar mass of PbCl2 for this.

.0525 moles PbCl2 x (278.11g/ 1mol PbCl2) = 14.60 g PbCl2

Good luck~!!

And if you wanted to know how both equations fit together you can do this:

.105 moles KCl x (1 mole PbCl2 / 2 moles KCl ) x (278.11g PbCl2/ 1mol PbCl2) = 14.60

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Answer 2

The first thing I would do is convert your 70mL to liters. Then I would take that and use it to calculate moles of KCl from 1.5M (M=moles per L). After you have moles of KCl, Its a matter of looking at the stoichiometry. For every 1 mole of PbCl2, you have 2 moles of KCl. If you take your KClmole calculation, 2:1 ratio, you should end up with moles of PbCl2. All you have to do now is convert your moles to grams. I came up with 14.61g.