[danny411]

hi

Im kinda lost on these ones.

oxidization/reduction reaction

we are trying to find charge

thats n=a/b

0.6215 g sample of nickel was dissolved in 100.0 ml of fe 3+

afe3+ +bNi--->afe2+ +b Ni+n

A 15.00 ml sample of this solution was taken and titrated. If it took 26.48ml of .0200M Cr2072- to titrate the solution, what is the value of n for the reaction above.

6fe2+ +cr2072- +14H---2cr3+ +6fe3+ + 7h20

I got this:

n=a/b

fe 3+=.02648 x .0200M x 6 fe2+/1 x 1mol fe3+/1 mol fe 2+=3.18 x 10-3 mol fe3+

Ni= .6215/55.84(of the top of my head)=1.06 x 10-2 mol

1.06 x 10-2 mol x 15/100=1.59 x10-3 mol

n=3.18 x10-3/1.59 x 10-3 =2

thanks again

[cliverlong]

hi

Im kinda lost on these ones.

oxidization/reduction reaction

we are trying to find charge

thats n=a/b

Well, I'm lost to.

What do n, a and b mean in your equation?

You can't quote and use an equation if you don't know or explain what the variables are.

afe3+ +bNi--->afe2+ +b Ni+n

Your equation is almost impossible to read and make sense of

Please use the buttons above the text area to create subscript and superscript. They are incredibly easy to use.

Can you use the correct symbols for chemical elements? If you don't it makes your question very difficult to read and understand.

Do you mean

aFe³⁺ + bNi  aFe²⁺ + bNi⁺ⁿ

?

If you can make the updates to the formatting in your question your problem may make more sense.

Can you give your question a more meaningful title e.g. "Oxidation question" rather than "can you check my stuff"? Are you asking me to look through your bag?


Clive

[danny411]

thanks for fixing the equation up. I didn't know you can do that

aFe³⁺ + bNi  right arrow aFe⁺² + bNi⁺ⁿ

n=charge on copper that is produced

a and b are the moles

the formula is n=a/b

the name of this unit is oxidization/reduction reactions.