pb(ch3coo)2 + h2s → pbs + ch3cooh (need to balance equation)

how many grams of pbs is produced when 5.00g of h2s is reacted with an excess (unlimited) supply of pb(ch3coo)2?

 Mass of PbS = 9.26 g

Explanation:

Data Given :

mass of H₂S = 5.00 g

mass of PbS = ?

Reaction Given:

                Pb(CH₃COO)₂ + H₂S → PbS + CH₃COOH

Solution:

Balance the equation:

                      Pb(CH₃COO)₂ + H₂S → PbS + 2CH₃COOH

Now Look for the number of moles of H₂S and PbS meta

                    Pb(CH₃COO)₂ +  H₂S  →  PbS  +  2CH₃COOH

                                                1 mol          1 mol

So,

1 mole of H₂S combine with excess (unlimited) supply of Pb(CH₃COO)₂ and produce 1 moles of PbS

Now Convert moles to mass for which we have to molar masses of  H₂S and PbS

Molar mass of H₂S = 2 + 32 = 43 g/mol

Molar mass of PbS  = 31 + 32 = 63 g/mol

                    Pb(CH₃COO)₂ +       H₂S       →            PbS      +  2CH₃COOH

                                                1 mol (34 g/mol)          1 mol (63 g/mol)

                                                     34 g                              63 g

So,

34 g of H₂S  produces 63 grams of PbS.

Now

What mass of silver is produced from 5 g of H₂S  

Apply unity formula

                          34 g of H₂S   ≅ 63 g of PbS

                          5 g of H₂S   ≅ X g of PbS

By doing cross multiplication

                     Mass of PbS = 63 g x 5 g / 34 g

                     Mass of PbS = 9.26 g