1948 United States presidential election in Iowa
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.
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All 10 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
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 County Results
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| Elections in Iowa | ||||||||||
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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 | % | |
|---|---|---|---|---|
| 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% | ||
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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.
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