1948 United States Presidential Election In Iowa
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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 representatives, or electors, to the United States Electoral College, Electoral College, who voted for President of the United States, president and Vice President of the United States, vice president. Iowa was won by incumbent Democratic Party (United States), Democratic President Harry S. Truman of neighbouring Missouri, running with United States Senate, Senator Alben W. Barkley, with 50.31% of the popular vote, against Governor of New York, Governor Thomas E. Dewey, Thomas Dewey (Republican Party (United States), R–New York (state), New York), running with Governor of California, Governor Earl Warren, with 47.58% of the popular vote. Results Results by county See also * United States presidential elections in Iowa References

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Iowa
Iowa () is a state in the Midwestern region of the United States, bordered by the Mississippi River to the east and the Missouri River and Big Sioux River to the west. It is bordered by six states: Wisconsin to the northeast, Illinois to the east and southeast, Missouri to the south, Nebraska to the west, South Dakota to the northwest, and Minnesota to the north. During the 18th and early 19th centuries, Iowa was a part of French Louisiana and Spanish Louisiana; its state flag is patterned after the flag of France. After the Louisiana Purchase, people laid the foundation for an agriculture-based economy in the heart of the Corn Belt. In the latter half of the 20th century, Iowa's agricultural economy transitioned to a diversified economy of advanced manufacturing, processing, financial services, information technology, biotechnology, and green energy production. Iowa is the 26th most extensive in total area and the 31st most populous of the 50 U.S. states, with a populat ...
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