If 10 g of Ag reacts with 1 g of sulphur, the amount of Ag2S formed will be [Atomic weight of Ag = 108, S = 32] ?

Explanation :  2  Ag +  S  →  Ag2S2  ×  108  g  of Ag reacts with 32  g of sulphur10 g  of Ag  reacts  with 32216 ×  10=320216 > 1 gIt means ‘S’ is limiting reagent32 g of S reacts to form 216 + 32 = 248 g of Ag2S1 g of S reacts to form 24832=7.75  gAlternative method :  neq of Ag = 10108=0.0925;neq of S = 116=0.0625 neq  = number  of  equivalentsSince  neq of S is  less than neq of Ag⇒  0.0625 eq of Ag will  react  with  0.0625  eq  of  S to  form0.0625  eq  of Ag2SHence,  amount of Ag2S  =  neq ×  Eq.  wt.  of  Ag2S=0.0626  ×  124=7.75g