1. Write the Balanced Chemical Equation

The reaction between nickel(III) sulfide (Ni₂S₃) and copper(I) nitrate (CuNO₃) produces copper(I) sulfide (Cu₂S) and nickel(II) nitrate (Ni(NO₃)₂). Here's the balanced equation:

Ni₂S₃ + 6 CuNO₃ → 3 Cu₂S + 2 Ni(NO₃)₂

2. Calculate the Moles of Each Reactant

* Moles of Ni₂S₃:

- Molar mass of Ni₂S₃ = (2 * 58.69 g/mol Ni) + (3 * 32.07 g/mol S) = 214.47 g/mol

- Moles of Ni₂S₃ = 42.35 g / 214.47 g/mol = 0.197 mol

* Moles of CuNO₃:

- Molar mass of CuNO₃ = 63.55 g/mol Cu + 14.01 g/mol N + 3 * 16.00 g/mol O = 125.56 g/mol

- Moles of CuNO₃ = 66.01 g / 125.56 g/mol = 0.525 mol

3. Determine the Limiting Reactant

The limiting reactant is the one that gets used up first and limits the amount of product that can be formed. To find this:

* Compare the mole ratio of the reactants to the stoichiometric ratio in the balanced equation. The balanced equation shows a 1:6 mole ratio of Ni₂S₃ to CuNO₃.

* Calculate the required moles of CuNO₃ to react completely with the given moles of Ni₂S₃:

- 0.197 mol Ni₂S₃ * (6 mol CuNO₃ / 1 mol Ni₂S₃) = 1.182 mol CuNO₃

* Since we only have 0.525 mol of CuNO₃, it is the limiting reactant.

4. Calculate the Theoretical Yield of Cu₂S

* Use the mole ratio from the balanced equation to determine the moles of Cu₂S formed from the limiting reactant (CuNO₃):

- 0.525 mol CuNO₃ * (3 mol Cu₂S / 6 mol CuNO₃) = 0.2625 mol Cu₂S

* Convert moles of Cu₂S to grams:

- Molar mass of Cu₂S = 2 * 63.55 g/mol Cu + 32.07 g/mol S = 159.17 g/mol

- Mass of Cu₂S = 0.2625 mol * 159.17 g/mol = 41.77 g

Therefore, the maximum amount of Cu₂S that can be formed is 41.77 g.