1966 Iowa gubernatorial election explained

Election Name:1966 Iowa gubernatorial election
Country:Iowa
Flag Image:Flag of Iowa (variant).svg
Type:Presidential
Ongoing:no
Previous Election:1964 Iowa gubernatorial election
Previous Year:1964
Next Election:1968 Iowa gubernatorial election
Next Year:1968
Election Date:November 8, 1966
Image1:File:Harold Hughes, US Senator.jpg
Nominee1:Harold Hughes
Party1:Democratic Party (United States)
Popular Vote1:494,259
Percentage1:55.34%
Nominee2:William G. Murray
Party2:Republican Party (United States)
Popular Vote2:394,518
Percentage2:44.17%
Map Size:240px
Governor
Before Election:Harold Hughes
Before Party:Democratic Party (United States)
After Election:Harold Hughes
After Party:Democratic Party (United States)

The 1966 Iowa gubernatorial election was held on November 8, 1966. Incumbent Democrat Harold Hughes defeated Republican nominee William G. Murray with 55.34% of the vote.

Primary elections

Primary elections were held on September 6, 1966.[1]

Democratic primary

Candidates

Results

Republican primary

Candidates

Results

General election

Candidates

Major party candidates

Other candidates

Results

Notes and References

  1. Web site: Summary of Official Canvass of Votes Cast in Iowa Primary Election . . 1966 . March 20, 2020.