If you have 215.8g of silver and 32.1g of sulfide how much silver sulfide do you have?

215.8 g Ag x 1 mol Ag/196.9 = 1.096 moles Ag32.1 g S x 1 mol S/32 g = 1 moles

Since Ag and S combine in a ratio of 2:1 (2 Ag to 1 S), the most Ag2S you can form is limited by moles of Ag. 1.096 moles Ag will react with 0.548 moles S ==> Ag2S. So the rest of the S will be left over and not used because there isn't enough Ag. Thus, the mass of resultant salt would be 197 g + 17.5 g = 214.5 g.