7 / 339.6 = 0.055...Moles of HCl used = 13.......
...Moles of Sb2S3 used = 18.7 / 339.6 = 0.055
...Moles of HCl used = 13.6 / 36.5 = 0.373
...Here, the molar ratio of Sb2S3 and HCl is:
...0.373:0.055
...When you compare the ratios, 1:6 is less than 0.373:0.055,
...therefore HCl used is limited.
...Calculate how much SbCl3 formed using moles of HCl used.
Using the moles that Pyae calculated, 0.055 mol Sb2S3 requires six times as many moles of HCl, which is 0.330 mol, and there are 0.373 moles of HCl. There's the answer, there is an excess of HCl. With all due respect, it's very difficult to follow the "logic" of all of the other ratios that Pyae uses. Clearly, there is a better way.
Yup, you have to consider limiting reagent. First, make sure the rxn is balanced:
Sb2S3 + 6HCl-->2SbCl3 + 3H2S
So this reaction requires 1 mol of Sb2S3 and 6 mol of HCl. Mole to mole ratio is:
1:6
Next find out the moles of reactants with given weights to compare the mole ratio to find out which one is limiting reagent.
Moles of Sb2S3 used = 18.7 / 339.6 = 0.055
Moles of HCl used = 13.6 / 36.5 = 0.373
Here, the molar ratio of Sb2S3 and HCl is:
0.373:0.055
When you compare the ratios, 1:6 is less than 0.373:0.055, therefore HCl used is limited. Calculate how much SbCl3 formed using moles of HCl used.
0.373 mol HCl x (2 mol of SbCl3 / 6 mol of HCl) x (228.13 g SbCl3 / mol SbCl3) = 28.36g SbCl3 formed.
