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, your calculations are correct.
To find the amount of reactant left, you first need to determine the moles of FeCl2 formed from each reactant.
For FeS, you are given that you have 0.223 moles. Using the stoichiometry of the balanced equation, you can convert moles of FeS to moles of FeCl2:
0.223 moles FeS × (1 mole FeCl2/1 mole FeS) = 0.223 moles FeCl2
For HCl, you are given that you have 0.652 moles. Again, using the stoichiometry of the balanced equation, you can convert moles of HCl to moles of FeCl2:
0.652 moles HCl × (1 mole FeCl2/2 moles HCl) = 0.326 moles FeCl2
Next, you calculate the percentage of HCl used, by dividing the moles of FeCl2 formed from HCl by the total moles of HCl:
(0.326 moles FeCl2 / 0.652 moles HCl) × 100 = 50% used HCl
Since the stoichiometry of the balanced equation tells us that 2 moles of HCl are required to form 1 mole of FeCl2, only half of the HCl has been used. Therefore, we can calculate the amount of HCl remaining by multiplying the percentage remaining (100% - 50% = 50%) by the initial moles of HCl:
50% remaining × (0.652 moles HCl) = 0.326 moles HCl remaining
So, you have correctly determined that 0.326 moles of HCl are remaining.