[oksanah]

Does anyone know...
How do you calculate percent yield of SnI2 in a two-part reaction...

Sn(S) + 2HCl-(L) → SnCl2(S) + H2+(G)
0.5463g Sn and 15mL of Hcl were added to get SnCl2. Then...

SnCl2(S) + 2KI-(L) → SnI2(S) +2KCl(S)
to the SnCl2, 2mL of KI were added.

I think that you need to figure out the theoretical yield of SnCl2, but then I don't understand what to do with the second reaction? Do you do another theoretical yield with the KI?

[Borek]

Yes, do the theoretical twice.

Note that you can do it in one step - 1 mole of Sn gives 1 mole of SnI₂, no matter what intermediate steps are.