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