Ferrous sulfate (like SMB) is selective for gold, it is not like iron metal that will cement other metals out of solution.
When trying to dissolve gold and make a solution of gold chloride, with HCl acid alone, the gold is far below hydrogen in the reactivity series, so it will not displace hydrogen from the HCl acid, the acid has basically no effect on the gold.
HCl is not a strong enough oxidizing acid for gold.
Nitric acid is a highly oxidizing acid, but it alone has virtually no effect on the gold, gold will not make nitrates.
To dissolve the gold we have to take electrons from the gold atoms, we need an acid to do this, we need a powerful oxidizing acid, and HCl with some nitric acid is an acid with the oxidizing capability to remove the electrons from the gold atoms.
Oxidize means to loose or take electrons.
Reduce means to gain or give back electrons.
when something is oxidized something has to be reduced.
when using acid as an oxidizing agent (acid) to oxidize metal, the acid is reduced as the metal is oxidized, saying this another way, as acid takes electrons from your metal the acid is reduced to a salt of that acid.
metal atoms + acid = ionic metal salt of that metal and acid
Copper metal + nitric acid = copper nitrate
The copper atom is oxidized to make copper ions, the nitric is reduced to nitrates.
The copper gave up electrons to the acid, the acid took electrons from the copper.
Example that hopefully will help you understand:
Now after dissolving the gold we have a dissolved salt of gold chloride, if there is no free nitric involved it is fairly easy to give back the electrons to gold ions so gold atoms will form again, with a reducing agent (like SMB, ferrous sulfate, or another metal like copper...).
But with free nitric acid, the gold that you are trying to reduce (give back electrons to with your reducing agent), is being oxidized back to gold ions by the aqua regia (HCL and nitric acid), or as gold is trying to gain electrons from your reducing agent your free nitric is oxidizing the gold right back into solution, as gold gives up the chloride some of the hydrogen from the free HNO3 forms back HCl in solution, that with the nitric acid in solution (basically reforming aqua regia to dissolve the fine gold powder basically as soon as it try's to form or precipitate.
Lets say we dissolve gold powder in a little HCl and just enough nitric, and use up the excess nitric acid, forming a gold chloride with no free acid. Now we can easily reduce the gold ions back to a metal powder of gold by adding a reducing agent. Say we use copper in this example, here the copper will be oxidized to copper ions of copper chloride, and the gold gaining the electrons from the copper will form a gold powder, once all of the gold has cemented, and as long as there were no free acid the copper would stop being dissolved (or would give up no more electron to the gold or the solution.
Now say we take this solution and remove the copper bar, but add a little a gram size bb of copper metal, with nothing to oxidize the copper it would just sit there with our gold powders precipitated in the bottom of the vessel of copper chloride.
Now what do you think would happen if you added some nitric acid to this salt solution of copper chloride and gold powders with the button of copper metal?
The HNO3 with the chloride salt would again form aqua regia strong enough to dissolve the copper, and strong enough to dissolve the gold powder again back into solution.
And if you had added enough excess acid you may have to add a lot more copper to cement your gold back out of solution, as the copper must consume the excess nitric before it will be able reduce the gold from solution, and displace your gold back out of solution.
