A current is passed through a Ga(NO3)3 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?

Question

supie · Answer

Choices would help...

supie · Answer

idk much about this but I can try....
$$Ga^3+ + 3e \ \ ➡\  Ga$$
A unit electric charge (faraday) and so $$1\ faraday= 96500C$$

supie · Answer

We have to do the value of 1 faraday multiplied by three $$3\ faraday = 3* 96500C = 289500C$$

supie · Answer

According to $$Ga^3++3e \  ➡ Ga$$$$70g/Ga$$ was deposited by 289500C meaning that for this situation $$6.3g/Ga$$ will now be deposited $$(6.3 x 289500)/6.3 = 26055C$$

supie · Answer

Now we calculate the `current flow` $$t = 1.5\ hours = 1.5 * 3600 = 5400\ seconds$$

supie · Answer

$$Q = 26055C$$$$Q = It$$$$I\ =\ Q/t$$$$I\ =\ 26055/5400$$ $$I \ =\  ?$$