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Schach [20]
2 years ago
14

How many amps are required to produce 75. 8 g of iron metal from a solution of aqueous iron(iii)chloride in 6. 75 hours?

Chemistry
1 answer:
Shalnov [3]2 years ago
7 0

The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

            Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

             I is the current passed

              t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

Learn more about Law of Equivalence here, brainly.com/question/13104984

#SPJ4

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It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.

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HOW TO CALCULATE MOLAR MASS:

The molar mass of a compound can be calculated by summing the atomic masses of its constituent elements.

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