A current is passed through a solution for 1.5 hours, and after this time period the mass of metal produced was 6.3 grams. What is the current, in amperes, that is required to produce such an amount of gallium

4.83A

Explanation:

The following were obtained from the question :

Mass of gallium deposited = 6.3g

Time = 1.5hrs = 1.5 x 60 x 60 = 5400secs

Gallium is a trivalent metal. This implies that it requires 3 faraday to deposit 1mole of gallium according to the equation

Ga^3+ + 3e —> Ga

Imole gallium = 70g

3 faraday = 3 x 96500C = 289500C

From the equation,

70g of gallium is deposited 289500C.

Therefore, 6.3g of gallium will be deposited by = (6.3 x 289500)/70

= 26055C.

Now we can find the current as follows

Q = It

I = Q/t

I = 26055/5400

I = 4.83A

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g