How many grams of nickel can be produced by passing a current of 15.0A through aqueous nickel(II) chloride for 80.0 minutes?
I'm not sure if I'm on the right path...
All I know is that
1 coloumb= amperes x seconds
1F=96485c/eq
eq=molar mass/charge
Since MM of Ni= 58.69g/mol and the charge is 2, eq= 29.345
If the current is 15A, and it runs for 80 minutes, then 80*60*15=72000C
Where do I go from here?
you're almost there.
29.93 g requires 96,485 C.
? g can we obtain from 72,000.
29.93 x (72,000/96,485) =
To determine the number of moles of nickel produced, you can use the following equation:
moles of nickel = (charge in coulombs / Faraday's constant) / equivalent weight of nickel
The charge in coulombs is 72000C, which was calculated correctly.
Faraday's constant, denoted by F, is equal to 96485 C/mol.
The equivalent weight of nickel, denoted by eq, is 29.345 g/eq, which you calculated correctly.
To calculate the number of moles of nickel:
moles of nickel = (72000C / 96485 C/mol) / 29.345 g/eq
Now, you can calculate the number of grams of nickel using the molar mass of nickel:
grams of nickel = moles of nickel * molar mass of nickel
The molar mass of nickel, denoted by MM, is 58.69 g/mol, as you correctly mentioned.
Hope this helps!
To calculate the number of grams of nickel produced, you need to use the equation:
grams of nickel = (charge * moles of nickel * molar mass of nickel) / Faraday's constant
First, let's calculate the moles of nickel:
moles of nickel = (charge * eq) / Faraday's constant
Given that the charge is 2, eq is 29.345, and Faraday's constant is 96485 C/eq, we can substitute and calculate:
moles of nickel = (2 * 29.345) / 96485 = 0.0000606686 moles
Next, let's calculate the grams of nickel:
grams of nickel = (0.0000606686 moles * 58.69 g/mol) / 1
Multiplying the moles by the molar mass of nickel, we get:
grams of nickel = 0.0000606686 * 58.69 = 0.003556607 g
Therefore, passing a current of 15.0 A through aqueous nickel(II) chloride for 80.0 minutes would produce approximately 0.0036 grams of nickel.