1948 United States presidential election in Iowa
From Wikipedia, the free encyclopedia
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.
Main article: 1948 United States presidential election
Quick Facts All 10 Iowa votes to the Electoral College, Nominee ...
| ||||||||||||||||||||||||||
All 10 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| ||||||||||||||||||||||||||
County Results
| ||||||||||||||||||||||||||
|
Close
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 Dewey (R–New York), running with Governor Earl Warren, with 47.58% of the popular vote.[3][4]