32.5 g of zinc interacts with 24 g of sulfur, find the mass of zinc sulfide, which thing is in excess?

Given:
m (Zn) = 32.5 g
m (S) = 24 g

To find:
m (ZnS) -?

Decision:
1) Zn + S => ZnS;
2) n (Zn) = m (Zn) / M (Zn) = 32.5 / 65 = 0.5 mol;
3) n (S) = m (S) / M (S) = 24/32 = 0.75 mol;
4) Sulfur in excess;
5) n (ZnS) = n (Zn) = 0.5 mol;
6) m (ZnS) = n (ZnS) * M (ZnS) = 0.5 * 97 = 48.5 g.

Answer: The mass of ZnS is 48.5 g.