Robert S. answered • 04/10/21
Patient PhD Chemist with 40 Years of R&D And Teaching
Hi, again, Lisa,
You are correct. You have more than one question. I'll take you through this one as I did another. It might be helpful if you could leave a comment on which part of the process is most confusing. It would help to know that when tutors respond.
The first step is to balance the equation.
Mg + CuCl2 🡪 MgCl2 + Cu
Hmmmm. I don't see anything wrong with the existing coefficients of "1" for each molecule. It is balanced already. And complete. Perhaps you got this far. If so, thanks, and good job.
We are given 50.6 grams of Mg (3 sig figs). Convert that to moles by dividing by Mg's molar mass of 24.3g/mole. That gives us 2.08 moles of Mg. If we assume there is plenty of copper chloride to consume all of the Mg, we'll get 2.08 moles of both the Cu and the MgCl2 . Then, convert those molar amounts into grams by multipkying by their molar masses:
Since the quetion onky asks for Cu, we'll do that first:
(2.08 moles Cu)*(63.546 grams/mole Cu) = 132 grams Cu
Wow. We started with 50.6 grams of Mg and wound up with 132 grams of copper? Let me in on this deal.
Well, maybe not. If, for fun, we see how much CuCl2 is required, the mass comes to 280 grams. The biggest impact is the difference in the atomic weights (molar masses) of Cu (63.5g/mole) and Mg (24.3mg/mole).
Also, for fun, we can determine the mass of MgCl2 produced, which is 198 grams.
Reactants: 50.6 g Mg and 280 grams CuCl2 equals 330.6 grams
Products: 198 grams MgCl2 and 132 grams Cu equals 330.5 grams total
Close enough, since I used 3 sig figs in the calculations.
I hope this helps,
Bob
