1904 United States presidential election in Iowa
Appearance
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Elections in Iowa |
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The 1904 United States presidential election in Iowa took place on November 8, 1904. All contemporary 45 states were part of the 1904 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 Theodore Roosevelt of New York and his running mate Charles W. Fairbanks of Indiana.
Results
United States presidential election in Iowa, 1904[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | Theodore Roosevelt | 308,158 | 63.39% | 13 | |
Democratic | Alton B. Parker | 149,276 | 30.71% | 0 | |
Social Democratic | Eugene Debs | 14,849 | 3.05% | 0 | |
Prohibition | Silas C. Swallow | 11,603 | 2.39% | 0 | |
Populist | Thomas E. Watson | 2,207 | 0.45% | 0 | |
Totals | 486,093 | 100.00% | 13 | ||
Voter turnout | — |
References
- ^ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Iowa