Can someone clarify this chem question

For the reaction
? Sb2S3 + ?HCl-->? SbCl3 + ?H2S ,
how many grams of SbCl3 (228.1 g/mol)
could be formed from 18.7 grams of Sb2S3
(339.6 g/mol) and 13.6 grams of HCl
(36.5 g/mol)?
Answer in units of grams.

Do they want me to consider limiting reactants? Or just come up with the ratios and moles myself. Because I did use limiting reactants to solve and it said I got it wrong, but only took off a point which means I was close.

This is a limiting reactant problem. Since the problem is asking for the mass of one of the products, the simplest solution is to compute the mass of that product using first one of the reactants and then the other. The actual mass of the product will be the smaller of the two.

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

18.7 g Sb2S3 x (1 mol Sb2S3 / 339.6 g) x (2 mol SbCl3 / 1 mol Sb2S3) x (228.1 g SbCl3 / 1 mol SbCl3) = 25.1 g SbCl3

13.6 g HCl x (1 mol HCl / 36.5g HCl) x (2 mol SbCl3 / 6 mol HCl) x (228.1 g SbCl3 / 1 mol SbCl3) = 28.3 g SbCl3

Since the smaller amount of product is 25.1 grams, that will be the mass of SbCl3 produced. And from this we can see that Sb2S3 is the limiting reactant.

================ Follow up ================

When specifically asked for the limiting reactant and not for a mass of product, you can do the following. Pick either one, it makes no difference, and determine the amount of the other reactant needed to react. Then you can tell if there is an excess or not of the second reactant. For example....

18.7 g Sb2S3 x (1 mol Sb2S3 / 339.6 g) x (6 mol HCl / 1 mol Sb2S3) x (36.5 g HCl / 1 mol HCl) = 12.1 g HCl

This means that when all of the 18.7 grams of of Sb2S3 have reacted, 12.1 grams of HCl will have reacted, but since you have 13.6 grams of HCl, the HCl is in excess, and Sb2S3 is the limiting reactant. This is the same conclusion that we reached in the example above.

Next we need to find out where Pyae's answer went astray. He wrote:

...Moles of Sb2S3 used = 18.7 / 339.6 = 0.055
...Moles of HCl used = 13.6 / 36.5