1880 United States presidential election in Iowa
The 1880 United States presidential election in Iowa took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.
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Elections in Iowa | ||||||||||
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Iowa voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the by a margin of 24.19%.
With 10.02% of the popular vote, Iowa would prove to be Greenback Party candidate James B. Weaver's second strongest state after Texas.[1]
Results
1880 United States presidential election in Iowa[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | James Abram Garfield of Ohio | Chester Alan Arthur of New York | 183,904 | 56.99% | 11 | 100.00% | ||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 105,845 | 32.80% | 0 | 0.00% | ||
Greenback | James Baird Weaver of Iowa | Barzillai Jefferson Chambers of Texas | 32,327 | 10.02% | 0 | 0.00% | ||
Prohibition | Neal Dow of Maine | Henry Adams Thompson of Ohio | 592 | 0.18 | 0 | 0.00% | ||
Total | 322,668 | 100.00% | 11 | 100.00% |
Notes
References
- "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1880 Presidential General Election Results - Iowa".
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