1867 Iowa gubernatorial election
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The 1867 Iowa gubernatorial election was held on October 8, 1867. Republican nominee Samuel Merrill defeated Democratic nominee Charles Mason with 58.88% of the vote.
General election
Candidates
- Samuel Merrill, Republican
- Charles Mason, Democratic
Results
Party | Candidate | Votes | % | ±% | |
---|---|---|---|---|---|
Republican | Samuel Merrill | 90,204 | 58.88% | ||
Democratic | Charles Mason | 62,966 | 41.10% | ||
Majority | 27,238 | ||||
Turnout | |||||
Republican hold | Swing |
References
- ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved June 5, 2021.
Categories:
- Articles with short description
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- Use mdy dates from September 2023
- Iowa gubernatorial elections
- 1867 United States gubernatorial elections
- 1867 Iowa elections
- October 1867 events
- United States gubernatorial elections in the 1860s
- 1867 in Iowa
- 1860s in Iowa
- Government of Iowa
- 1867 elections
- 1867 elections in North America
- 1867 elections in the United States