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1948 United States presidential election in Iowa

From Wikipedia, the free encyclopedia

1948 United States presidential election 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.

Quick facts All 10 Iowa votes to the Electoral College, Nominee ...

Iowa was won by incumbent Democratic President Harry S. Truman of neighbouring Missouri, running with Senator Alben W. Barkley, with 50.31% of the popular vote, against Governor Thomas E. Dewey (RNew York), running with Governor Earl Warren, with 47.58% of the popular vote.[3][4]

Defying the expectations of a Dewey sweep in a state he won four years earlier, in an unexpected twist, Truman would manage to flip the state from Republican to Democratic, a swing of roughly 6%.

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Results

More information Party, Candidate ...

Results by county

More information County, Harry S. Truman Democratic ...

Counties that flipped from Democratic to Republican

Counties that flipped from Republican to Democratic

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See also

References

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