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Here is the full lab:

Lab: Electrolytic Cells

Purpose: The purpose of this experiment is to test the method of stoichiometry in cells.

Materials:
Balance
Steel can
Tin electrode
Power Source
Wire lead (x2)
tin(II) chloride solution (3.25 M)
Timer
Large beaker

Procedure:
Take the mass of the steel can and record your observations.

I get 117.34 grams

Place the steel can in the large beaker.

Pour the solution of tin(II) chloride into the beaker so it surrounds (but does not go in) the steel can.

Place the tin electrode into the beaker.

Attach the wires to the steel can and the tin electrode.

Connect the wires to the proper terminus on the power source.

Set the timer to 6.00 minutes and start the timer.

The power source will come on automatically and will turn off automatically when time runs out.

Take the mass of the steel can.Assume it is completly dry when you do so.

I get 118.05 grams

Analysis:
What is the mass of the tin produced?

It is 0.71 grams

What is the theoretical mass of tin that should have been produced?

Moles SnCl2 = 0.71 g / 189.60 g/mol =0.003744
The ratio is 1 : 1
Moles Sn produces = 0.03744
Mass Sn = 0.03744 mol x 118.69 g/mol = 0.444 grams

5 answers

What is the mass of the tin produced?

It is 0.71 grams OK here.

What is the theoretical mass of tin that should have been produced?
Theoretical mass means what is the theoretical mass based upon the electrolysis that took place.
ampere x seconds = coulombs.
How many mols is that? 95,485 coulombs will use 1 mol electrons. So ??coulombs/96,485 = mols.What is the equation for the deposition of Sn? How many electrons are involed in that deposition. Convert to mols Sn, then to grams Sn.

Moles SnCl2 = 0.71 g / 189.60 g/mol =0.003744
The ratio is 1 : 1
Moles Sn produces = 0.03744
Mass Sn = 0.03744 mol x 118.69 g/mol = 0.444 grams
Note that I made a typo above. It should be 96,485 coulombs and not 95, 485.
Do you know a website where I can find this equations? (for disposition of tin)? I have searched through many webpages but got no information
equation*
Answered in your later post.