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MArishka [77]
3 years ago
8

A current of 0.15 A is passed through an aqueous solution of K2PtCl4. How long will it take to deposit 1.00 g Pt(s) (M = 195.1)?

Chemistry
1 answer:
koban [17]3 years ago
7 0

First calculate the electric charge used to deposit 1.0 g Pt

C = (1.0 g Pt) (1 mol Pt / 195.1 g Pt) ( 2 mol e / 1 mol Pt) ( 96485 C / 1 mol e)

C = 989.08 C

C = It

Where I is the current

T is the time

T = C / i

T = 989.08 C / 0.15 A

T = 6593.88 s

T = 1.83 hrs

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                    No the substance is not water.

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Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

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According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

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Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

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                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

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