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1896 United States presidential election in Kansas
From Wikipedia, the free encyclopedia
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The 1896 United States presidential election in Kansas took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Kansas voters chose ten electors to the Electoral College, which selected the president and vice president.
Kansas was won by the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. The Republican nominees, former Governor of Ohio William McKinley and his running mate Garret Hobart of New Jersey. Bryan won the state by a narrow margin of 3.69%.
With his win in the state, Bryan became the first Democratic presidential candidate to win the state of Kansas. Bryan would later lose Kansas to McKinley four years later during their rematch and would later lose the state again to William Howard Taft in 1908. This remains the only election since Kansas statehood in which the Republican candidate won the presidency without Kansas, or where Kansas voted more Democratic than Maryland.
This was the last election until 2020 in which a Democrat carried Johnson County with a majority. McKinley was also the last Republican to win the White House without carrying this traditionally Republican county until Donald Trump in 2024.
This was the first of six elections in which the Democratic nominee carried Kansas, traditionally a Republican bastion like the other Plains States. The others were 1912, 1916, 1932, 1936, and 1964.[1]
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Results
Results by county
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See also
Notes
References
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