United States presidential election in Iowa, 1900

United States presidential election in Iowa, 1900
November 6, 1900

 
Nominee William McKinley William J. Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai Stevenson I
Electoral vote 13 0
Popular vote 307,808 209,265
Percentage 58.04% 39.46%
President before election

William McKinley
Republican

 Elected President

William McKinley
Republican

The 1900 United States presidential election in Iowa took place on November 6, 1900. All contemporary 45 states were part of the 1900 United States presidential election. Iowa voters chose thirteen electors to the Electoral College, which selected the president and vice president.

Iowa was won by the Republican nominees, incumbent President William McKinley of Ohio and his running mate Theodore Roosevelt of New York.

Results

United States presidential election in Iowa, 1900[1]
Party Candidate Votes Percentage Electoral votes
Republican William McKinley 307,808 58.04% 13
Democratic William Jennings Bryan 209,265 39.46% 0
Prohibition John Woolley 9,502 1.79% 0
Social Democratic Eugene Debs 2,742 0.52% 0
Populist Wharton Barker 613 0.12% 0
Socialist Labor Joseph Maloney 259 0.05% 0
United Christian Jonah Leonard 166 0.03% 0
Totals 530,355 100.00% 13
Voter turnout —

References

  1. ↑ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Iowa

Notes

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