1880_United_States_presidential_election_in_Iowa

1880 United States presidential election in Iowa

1880 United States presidential election in Iowa

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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.

Quick Facts Nominee, Party ...

Iowa voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state 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

More information Party, Candidate ...

Results by county

More information County, James Abram Garfield Republican ...

See also

Notes

  1. In this county where Hancock ran third behind Garfield and Weaver, margin given is Garfield vote minus Weaver vote and percentage margin Garfield percentage minus Weaver percentage.
  2. In this county where Garfield ran third behind Hancock and Weaver, margin given is Hancock vote minus Weaver vote and percentage margin Hancock percentage minus Weaver percentage.

References

  1. "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. Géoelections; Popular Vote at the Presidential Election for 1880 (.xlsx file for €15)

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