United States presidential election in Iowa, 1896
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Elections in Iowa |
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The 1896 United States presidential election in Iowa took place on November 3, 1896. All contemporary 45 states were part of the 1896 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, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey.
Results
United States presidential election in Iowa, 1896[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | William McKinley | 289,293 | 55.47% | 13 | |
Democratic | William Jennings Bryan | 223,741 | 42.90% | 0 | |
National Democratic | John M. Palmer | 4,516 | 0.87% | 0 | |
Prohibition | Joshua Levering | 3,192 | 0.61% | 0 | |
Socialist Labor | Charles Matchett | 453 | 0.09% | 0 | |
National Prohibition | Charles Bentley | 352 | 0.07% | 0 | |
Totals | 521,547 | 100.00% | 13 | ||
Voter turnout | — |
References
- ↑ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Iowa
Notes
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Wikipedia.
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