Question

Be sure to answer all parts. Calculate the mass of each product formed when 683 g of silver sulfide reacts with excess hydrochloric acid: Ag2S(s)+HCaq) AgC+ H2S) unbalanced Mass of AgCl Mass of H2S:

Answer 1

The balanced equation is

Ag₂S (s) + 2 HCl (aq) -----> 2 AgCl (s) + H₂S (g)

Number of moles of Ag2S = 683 g / 247.8 g/mol = 2.76 mol

From the balanced equation we can say that

1 mole of Ag2S produces 2 mole of AgCl so

2.76 mole of Ag2S will produce

= 2.76 mole of Ag2S *(2 mole of AgCl / 1 mole of Ag2S)

= 5.52 mole of AgCl

mass of 1 mole of AgCl = 143.32 g

so the mass of 5.52 mole of AgCl = 791 g

Therefore, the mass of AgCl produced would be 791 g

From the balanced equation we can say that

1 mole of Ag2S produces 1 mole of H2S so

2.76 mole of Ag2S will produce

= 2.76 mole of Ag2S *(1 mole of H2S /1 mole of Ag2S)

= 2.76 mole of H2S

mass of 1 mole of H2S = 34.1 g

so the mass of 2.76 mole of H2S = 94.1 g

Therefore, the mass of H2S produced would be 94.1 g