I just need a little help with my chemistry.?

Question

It says that a student obtained the following data from a reaction between a zinc strip and a copper II sulfate solutioin. WHat mass of copper would be produced?

initial mass of zinc 15.42g
final mass of zinc 14.15g

Answer 1

Cu molecular mass = 63.546
Zn molecular mass = 65.409
The reaction is

CuSO4 + Zn -> ZnSO4 + Cu

One mole of Zn reacts with one mole of CuSO4 to produce 1 mole of Cu. The reaction used 15.42-14.15 = 1.27 g zinc, or

1.27/65.409 = 0.0194 moles of zinc; this must equal the moles of Cu,

0.0194*63.546 = 1.234 g of Cu

Answer 2

Write out the single replacement chemical reaction which is occurring…
The Zinc metal is higher on the activity series than Copper so the Zn will displace the Cu from solution, precipiting out Copper metal and leaving Zn+2 in solution.
Zn (s) + Cu+2 (aq) --> Cu (s) + Zn+2 (aq)
For every 1 mole of Zinc metal lost to solution, 1 mole of Copper metal is obtained.

In your experiment results, 1.27 grams of Zinc went into solution (= the difference in the mass of the Zinc strip in the initial and final case).
Dividing 1.27 grams by the molar mass of Zinc we can find the number of moles of Zn which went into solution.
1.27 grams / 65.41 g/mol = .0194 moles of Zinc

From the mole ratio in the above chemical reaction, this means that .0194 moles of Copper came out of solution. Multiplying the moles of Copper by the molar mass of Copper (63.55 g/mol), we can find that the mass of the Copper precipitate is 1.23 grams of Copper metal.
.0194 moles of Cu * 63.55 g/mol Cu = 1.23 grams Cu


http://www.webelements.com/webelements/elements/text/Zn/key.html
http://www.webelements.com/webelements/elements/text/Cu/key.html

Answer 3

.27 gms
As Zn is remaining CuSo4 is the limiting reagent.

Answer 4

Zn + CuSo4 → ZnSO4 + Cu
∆Zn = 1.27 g
∆Cu = 1.27*63.55/65.39
∆Cu = 1.23 g

Answer 5

well you have the expert here(mrjeffry3)
belive him.

Answer 6

0.27g