How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from `CaCl_(2)` (ii) 40.0 g of Al from molten `Al_(2)O_(3)`?

How much electricity in terms of faraday is required to produce (i) 20

1.	How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from `CaCl_(2)` (ii) 40.0 g of Al from molten `Al_(2)O_(3)`?
Answer» (i) `Cu^(2+)+2e^(-)toCa` Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity=2F`therefore20` g of Ca will require electricity=1F (ii) `Al^(3+)+3e^(-)toAl` . Thus 1 mol of Al, i.e., 27g of Al required electricity=3F `therefore40g` of Al will require electricity=`(3)/(27)xx40=4.44E`

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How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from `CaCl_(2)` (ii) 40.0 g of Al from molten `Al_(2)O_(3)`?

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