Calculate the mass of each product formed when 84.3 g of silver sulfide reacts with excess hydrochloric acid: ag2s(s) + hcl(aq) → agcl(s) + h2s(g)

              48.75 g of AgCl

               11.60 g of H₂S

Solution:

The Balance Chemical Equation is as follow,

                                     Ag₂S  +  HCl     →     AgCl  +  H₂S

Calculate amount of AgCl produced,

According to equation,

                247.8 g (1 mol) of Ag₂S produces  =  143.32 g (1 mol) of AgCl

So,

                         84.3 g of Ag₂S will produce  =  X g of AgCl

Solving for X,

                      X  =  (84.3 g × 143.32 g) ÷ 247.8 g

                      X  =  48.75 g of AgCl

Calculate amount of H₂S produced,

According to equation,

                247.8 g (1 mol) of Ag₂S produces  =  34.1 g (1 mol) of H₂S

So,

                         84.3 g of Ag₂S will produce  =  X g of H₂S

Solving for X,

                      X  =  (84.3 g × 34.1 g) ÷ 247.8 g

                      X  =  11.60 g of H₂S

i'm pretty sure it is letter c.) car