United States presidential election in Iowa, 1904

United States presidential election in Iowa, 1904
November 8, 1904

 
Nominee Theodore Roosevelt Alton B. Parker
Party Republican Democratic
Home state New York New York
Running mate Charles W. Fairbanks Henry G. Davis
Electoral vote 13 0
Popular vote 308,158 149,276
Percentage 63.39% 30.71%
President before election

Theodore Roosevelt
Republican

 Elected President

Theodore Roosevelt
Republican

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]
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

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

Notes

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