We are given the amount of </span><span>antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:
2.85 g Sb2S3 ( 1 mol / </span><span>339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) (</span> 228.13 g / mol ) = 11.48 g SbCl3