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Contents

   



(Top)
 


1 General election  



1.1  Candidates  





1.2  Results  







2 References  














1867 Iowa gubernatorial election







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From Wikipedia, the free encyclopedia
 


1867 Iowa gubernatorial election

← 1865 October 8, 1867 1869 →
 
Nominee Samuel Merrill Charles Mason
Party Republican Democratic
Popular vote 90,204 62,966
Percentage 58.88% 41.10%

County results
Merrill:      50-60%      60-70%      70-80%      80-90%      90-100%
Mason:      50-60%      60-70%
No Data/Votes:      


Governor before election

William M. Stone
Republican

Elected Governor

Samuel Merrill
Republican

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[edit]

Candidates[edit]

Results[edit]

1867 Iowa gubernatorial election[1]
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[edit]

  1. ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved June 5, 2021.

Retrieved from "https://en.wikipedia.org/w/index.php?title=1867_Iowa_gubernatorial_election&oldid=1233066974"

Categories: 
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
Hidden categories: 
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Use mdy dates from September 2023
 



This page was last edited on 7 July 2024, at 02:51 (UTC).

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