10g of Sb2S3 reacts with 7.3g of HCl. how much of the reactant in excess is left at the end?

Question

Ive worked out the limiting reactant which is Sb2S3. Ive calculated moles of Sb2S3. Do I multiply moles of Sb2S3 by 6 to get moles of HCl used? How do you work the mass in excess from there?

 

The equation:

Sb2S3 + 6HCl------- 2SbCl3 + 3H2S

Answer 1

Sb₂S₃ + 6HCl------- 2SbCl₃ + 3H₂S

 

You are correct that Sb2S3 is the limiting reactant.

Yes, you multiply moles of Sb2S3 by 6 to get moles of HCl used.

 

After you obtain moles of HCl used, you simply subtract that number from moles of HCl initially present, which you must have calculated already in order to know that it was not limiting.  You then convert to mass using molar mass HCl.

 

Does that answer your question?  You should get an answer of 0.86 g HCl in excess.(2 sig. figs.)