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

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.