1. Write the Balanced Chemical Equation:

The reaction between zinc (Zn) and sulfur (S) produces zinc sulfide (ZnS):

Zn + S → ZnS

2. Calculate the Molar Mass of Each Reactant:

* Zn: 65.38 g/mol

* S: 32.06 g/mol

3. Calculate Moles of Each Reactant:

* Moles of Zn = (25 g) / (65.38 g/mol) = 0.383 mol

* Moles of S = (300 g) / (32.06 g/mol) = 9.35 mol

4. Determine the Limiting Reactant:

The limiting reactant is the one that gets used up first, dictating the amount of product formed. To figure this out, we need to compare the mole ratios from the balanced equation:

* The balanced equation shows a 1:1 mole ratio between Zn and S. This means for every 1 mole of Zn, we need 1 mole of S to react completely.

Since we have 0.383 moles of Zn and 9.35 moles of S, we can see that we have significantly more sulfur than needed. Therefore, zinc (Zn) is the limiting reactant.

5. Calculate the Mass of Excess Reactant:

To find the mass of excess reactant (sulfur), we first need to determine how much sulfur is needed to react with the limiting reactant (zinc):

* Moles of S needed = 0.383 moles Zn * (1 mole S / 1 mole Zn) = 0.383 moles S

Now, we can calculate the mass of sulfur needed:

* Mass of S needed = 0.383 moles * 32.06 g/mol = 12.27 g

Finally, calculate the mass of excess sulfur:

* Mass of excess S = 300 g (initial) - 12.27 g (needed) = 287.73 g

Answer: The mass excess reactant is 287.73 g of sulfur.