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How do I calculate excess reagent?
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
A reaction mixture initially contains 0.236 mol FeS and 0.664 mol HCl.
Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?
My original thought was 0.096 moles because 0.664/2 - 0.236 is 0.096, but this isn't right. What am I doing wrong?
The answer and original values are given in moles.
3 Answers
- sioiengLv 73 years agoFavorite Answer
Initial number of moles of FeS = 0.236 mol
Initial number of moles of HCl = 0.664 mol
FeS(s) + 2HCl(aq) → FeCl₂(s) + H₂S(g)
Mole ratio FeS : HCl = 1 : 2
When 0.236 mol of FeS completely reacts, HCl needed = (0.236 mol) × 2 = 0.472 mol < 0.664 mol
Hence, HCl is in excess.
Number of moles of excess HCl = (0.664 – 0.472) mol = 0.192 mol
- busterwasmycatLv 73 years ago
It is the other way around. Calculate the smaller amount (in this case the mole FeS) and take the reaction equivalent (twice that amount of HCl is needed). Subtract that total consumption from the starting value for HCl.
You basically ended up with half of the actual residue (excess reagent) because you calculated how much more FeS would be required to use all the HCl, instead of how much HCl was left after using all the FeS.
