Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated naoh(aq). calculate the amount of ga(s) that can be deposited from a ga(iii) solution by a current of 0.610 a that flows for 60.0 min.

First, we need to determine the half reaction of gallium or the electrolysis reaction. It would be expressed as:

Ga3+ + 3e = Ga(s) 

Given the mass of current and time for the process of electrolysis, we can calculate the mass of solid gallium that is deposited from the total charge of the process. Then, from the relation 1 mol of electrons is equal to 96500 C and the relation of the electron and the solid gallium, we can determine the moles of Gallium. We calculate as follows:

q (charge) = It
              q = (0.610 A) (60.0 min) ( 60 s / 1 min )
              q = 2196 C

moles Gallium = 2196 C ( 1 mol e- / 96500 C ) ( 1 mol Ga / 3 mol e- ) = 7.585x10^-3 mol Ga
mass Ga =  7.585x10^-3 mol Ga ( 69.72 g / 1 mol ) = 0.53 g Gallium

Ibelieve your answer is true 

Compound with sulfer  is bright red