Huron, Indiana
From Wikipedia, the free encyclopedia
Huron, Indiana | |
|---|---|
| Coordinates: 38°43′16″N 86°40′15″W / 38.72111°N 86.67083°W | |
| Country | United States |
| State | Indiana |
| County | Lawrence |
| Township | Spice Valley |
| Elevation | 564 ft (172 m) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| ZIP code | 47446 |
| Area code | 812 |
| GNIS feature ID | 2830451[1] |
Huron is a census-designated place in Spice Valley Township, Lawrence County, in the U.S. state of Indiana.[1]
