Question

To a beaker containing 200 mL of a solution of CuSO4 was added 10.0g of magnesium metal. When the reaction was complete, a mixture of Mg and Cu having a mass of 14.50g remained in the beaker. Calculate the molarity of the original CuSO4 solution.

I understand that I need to get the original number of moles, and to get that, I'm confused. Can I divide the mass of the resultant (14.50g) into 2? But if I do that then reacted line (in a IRF table) doesnt work. Also is the equation CuSO4(aq) + Mg2(s) = Mg2(s) + Cu(s) + SO4(aq)?
The equation is
CuSO4 + Mg ==> MgSO4 + Cu
I don't know that dividing 14.5/2 or 2/14.5 will accomplish anything. Do you have ANYTHING else in the problem?

Related Questions

What volume of H2O should be added to 500 ML of a 1.0 mol/L CuSO4 solution to dilute it to 0.5 mol/L... I did an experiment were I added some water and KHP into a beaker and phenophalene [sp?]. It was a t... Tim adds a magnesium strip to a beaker containing hydrochloric acid. He notices that bubbles immedia... What is the correct symbol equation for the reaction of magnesium with copper sulphate? 1. Mg(s) +...