The reduction of iodine gas at a constant current of 1.50 amperes for 12.0 hours would result in how many grams of i–? (1 faraday = 96,500 coulombs)

85.2 grams.

Explanation:

Look up the relative atomic mass data on a modern periodic table.

The Faraday's Constant gives the size of the charge (in Coulombs) on one mole of electrons. Also, keep in mind that .

Charge supplied to the iodine:

How many moles of electrons does that much charge correspond to?

.

Consider the reduction half-equation for iodine:

.

Divide both sides by 2 to make sure that the coefficient in front of is equal to 1.

.

The ratio between the coefficient in front of and that of is equal to 1. In other words, while gaseous iodine is in excess, each mole of electrons will produce one mole of iodine ions.

of electrons will produce of iodine ions.

The mass of one mole of iodine ions is approximately the same as that of one mole of iodine atoms.

.

.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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