1884 United States presidential election in Iowa
The 1884 United States presidential election in Iowa took place on November 4, 1884. All contemporary 38 states were part of the 1884 United States presidential election. Iowa voters chose thirteen electors to the Electoral College, which selected the president and vice president. [1]
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Elections in Iowa | ||||||||||
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Iowa was won by Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 52.25% of the vote, against Grover Cleveland, the 28th governor of New York, (D–New York), running with the former governor of Indiana Thomas A. Hendricks, with 47.01% of the popular vote.[1]
The Prohibition party chose the 8th Governor of Kansas, John St. John and Maryland State Representative William Daniel, received 0.40% of the popular vote.
Results
1884 United States presidential election in Iowa | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | James G. Blaine | 197,089 | 52.25% | 13 | |
Democratic | Grover Cleveland | 177,316 | 47.01% | 0 | |
Prohibition | John St. John | 1,499 | 0.40% | 0 | |
No party | Write-ins | 1,297 | 0.34% | 0 |
References
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