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
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Given a mole each for iron and magnesium, the number of atoms of each element is equal. Iron has a greater mass due to its greater molecular weight. The correct statement among the choices is D.
a. mol O₂=0.5
b. volume O₂ = 25 cm³
c. i. the total volume of the two reactants = 75 cm³
c. ii. the volume of nitrogen dioxide formed = 50 cm³
<h3>Further explanation</h3>
Reaction
2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)
a.
mol NO = 1
From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :
b.
From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :
c.
i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³
ii. the volume of nitrogen dioxide formed :
mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³
