1948 United States presidential election in Iowa

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United States presidential election in Iowa, 1948

← 1944 November 2, 1948[1] 1952 →

All 10 Iowa votes to the Electoral College
  Harry S. Truman.jpg ThomasDewey.png
Nominee Harry S. Truman Thomas Dewey
Party Democratic Republican
Home state Missouri New York
Running mate Alben W. Barkley Earl Warren
Electoral vote 10 0
Popular vote 522,380 494,018
Percentage 50.3% 47.6%

President before election

Harry S. Truman
Democratic

Elected President

Harry S. Truman
Democratic

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 (DMissouri), running with Senator Alben W. Barkley, with 50.31% of the popular vote, against Governor Thomas Dewey (RNew York), running with Governor Earl Warren, with 47.58% of the popular vote.[3][4]

Results[edit]

United States presidential election in Iowa, 1948
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%

References[edit]

  1. ^ "United States Presidential election of 1948 - Encyclopædia Britannica". Retrieved October 26, 2017.
  2. ^ "1948 Election for the Forty-First Term (1949-53)". Retrieved October 26, 2017.
  3. ^ "1948 Presidential General Election Results - Iowa". Retrieved October 26, 2017.
  4. ^ "The American Presidency Project - Election of 1948". Retrieved October 26, 2017.