If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
maximum mass of ammonium sulfide that can be formed?

2 NH3 + H2S > (NH4)2S

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

Explanation:

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce g of (NH₄)₂S

25g of NH₃ will produce of (NH₄)₂S

                                     = 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

Answer & Explanation:

The balanced equation for the reaction of combustion of methane:

CH₄ + 2O₂   →  CO₂ + 2H₂O

    reactants   products

C       1                 1

H      4                 4

O      4                 4

Moles of Methane = mass ÷ molar mass

                              = 10 g ÷ 16.04 g/mol

                              = 0.6234 mol

Since the mole ratio of CH₄ to CO₂ based on the balance equation is 1:1, then for every mole of  CH₄ reacted, 1 mole of CO₂.

∴ since mole of  CH₄ = 0.6324 mol

then moles of  CO₂  = 0.6324 mol

⇒ Mass of  CO₂ produced = moles × molar mass

                                            = 0.6324 mol × 44.01 g/mol

                                            = 27.83 g

                                            ≈ 27 g (as the question predicts).

Law of Conservation of Mass:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.   According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.