The problem asked to find the amount (in moles) of reactant left.
.223moles FeS and .652moles of HCl
FeS+2HCl--> FeCl2+H2S
.223molFeS(1moleFeCl2/1moleFeS)=.223moleFeCl2
.652molHCl(1moleFeCl2/2molHCl)=.326moleFeCl2
.223/.326x100=68.4% used HCL
100-68.4=31.6% remaining
31.6%x(.652molHCl)=0.206mol HCl remaning
I was just wondering if I did it correctly.
Yes, 0.206 mol HCl remaining is exactly right; however you went around the barn to work the problem. Here is an easier way.
Let's say you've already determined that FeS is the limiting reagent.
So 0.223 mols FeS x (2 mols HCl/1 mol FeS) = 0.446 mols HCl used.
0.652 mols HCl initially - 0.446 mols HCl used = 0.206 mols HCl remaining.
The percentage steps are not necessary.