So I am going to react hydrochloric acid with tin and want to get a rough estimate of how much tin a given quantity of acid can dissolve.

Atomic weights: Sn = 119, Cl = 35.5, H = 1

100ml of 35% HCl contains 35ml of HCl or approximately 35g HCl, about 1 mole of chlorine.

In the reaction, 1 mole of tin reacts with 2 moles of chlorine. As we have only 1 mole of chlorine then the acid would react with half a mole of tin.
About 60g of tin.

Can anyone tell me if this about right?
It seems an awful lot of tin.

 

Does not easily react ...

At wt Sn is 119 ............HCl is 36 .........

So 119 gm Sn reacts with 72 gm HCl ........... but since acid is only 35% pure HCl , you need 200gm or ml (aprox) of acid

Or 60gm tin reacts with about 100gm acid ... but that's assuming all acid reacts , as the reaction proceeds acid will dilute and make things very slow ..

It's customary to use excess acid , then heat . The HCl and water left over will be driven off ..

 

OK, thanks. Given that it's 40 or so years since I did any such calculations I am so pleased I got it right. The plan is to deliberately leave some acid there so it is an acidic stannous chloride solution. So I can put in 30g of tin to get the desired result - but as I said it seems a lot of tin.