1940 United States presidential election in Iowa
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All 11 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
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| Elections in Iowa |
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The 1940 United States presidential election in Iowa took place on November 5, 1940, as part of the 1940 United States presidential election. Iowa voters chose eleven[2] representatives, or electors, to the Electoral College, who voted for president and vice president.
Iowa was won by Wendell Willkie (R–Indiana), running with Minority Leader Charles L. McNary, with 52.03% of the popular vote, against incumbent President Franklin D. Roosevelt (D–New York), running with Secretary of Agriculture Henry A. Wallace, with 47.62% of the popular vote.[3][4]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Wendell Willkie | 632,370 | 52.03% | |
| Democratic | Franklin D. Roosevelt (inc.) | 578,800 | 47.62% | |
| Write-in | 4,260 | 0.35% | ||
| Total votes | 1,215,430 | 100% | ||
References
- ^ "United States Presidential election of 1940 - Encyclopædia Britannica". Retrieved August 19, 2018.
- ^ "1940 Election for the Thirty-ninth Term (1941-45)". Retrieved August 19, 2018.
- ^ "1940 Presidential General Election Results - Iowa". Retrieved August 19, 2018.
- ^ "The American Presidency Project - Election of 1940". Retrieved August 19, 2018.
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