0

Guest

Courses New

How to calculate the no. of electrons required to deposit 40.5 g of Al

How to calculate the no. of electrons required to deposit 40.5 g of Al

Grade:12

1 Answers

varun
304 Points
6 years ago
HELLO NADIA,
 The reaction is,
Al3+        +            3e-   ----------------------->   Al
1mole                    3mole                            1mole
1mole                     3faradays                      1 mole
 
1 mole of Al = 27gm is deposited by 3 faraday of electricity
therefore,
40.5gm of Al is deposited by = 3/27 X 40.5F
 
                                             = 3/27 X 40.5* 96500
                                             = 434250 Coulombs

Think You Can Provide A Better Answer ?

Other Related Questions on Inorganic Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More