1948 United States presidential election in Iowa
Appearance
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All 10 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
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Elections in Iowa |
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The 1948 United States presidential election in Iowa took place on November 2, 1948, as part of the 1948 United States presidential election. Iowa voters chose ten[2] representatives, or electors, to the Electoral College, who voted for president and vice president.
Iowa was won by incumbent President Harry S. Truman (D–Missouri), running with Senator Alben W. Barkley, with 50.31% of the popular vote, against Governor Thomas Dewey (R–New York), running with Governor Earl Warren, with 47.58% of the popular vote.[3][4]
Results
Party | Candidate | Votes | % | |
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Democratic | Harry S. Truman (inc.) | 522,380 | 50.31% | |
Republican | Thomas Dewey | 494,018 | 47.58% | |
Progressive | Henry A. Wallace | 12,125 | 1.17% | |
Socialist Labor | Edward A. Teichert | 4,274 | 0.41% | |
Prohibition | Claude A. Watson | 3,382 | 0.33% | |
Write-in | 2,093 | 0.20% | ||
Total votes | 1,038,272 | 100% |
References
- ^ "United States Presidential election of 1948 - Encyclopædia Britannica". Retrieved October 26, 2017.
- ^ "1948 Election for the Forty-First Term (1949-53)". Retrieved October 26, 2017.
- ^ "1948 Presidential General Election Results - Iowa". Retrieved October 26, 2017.
- ^ "The American Presidency Project - Election of 1948". Retrieved October 26, 2017.