1880 United States presidential election in Missouri explained

See main article: 1880 United States presidential election.

Election Name:1880 United States presidential election in Missouri
Country:Missouri
Type:presidential
Ongoing:no
Previous Election:1876 United States presidential election in Missouri
Previous Year:1876
Next Election:1884 United States presidential election in Missouri
Next Year:1884
Election Date:November 2, 1880
Image1:WinfieldScottHancock2 (cropped 3x4).jpg
Nominee1:Winfield S. Hancock
Party1:Democratic Party (United States)
Home State1:Pennsylvania
Running Mate1:William H. English
Electoral Vote1:15
Popular Vote1:208,600
Percentage1:
Nominee2:James A. Garfield
Party2:Republican Party (United States)
Home State2:Ohio
Running Mate2:Chester A. Arthur
Electoral Vote2:0
Popular Vote2:153,647
Image3:James Weaver - Brady-Handy (cropped 3x4).jpg
Nominee3:James B. Weaver
Party3:Greenback Party
Home State3:Iowa
Running Mate3:Barzillai J. Chambers
Electoral Vote3:0
Popular Vote3:35,042
Percentage3:8.82%
Map Size:340px
President
Before Election:Rutherford B. Hayes
Before Party:Republican Party (United States)
After Election:James A. Garfield
After Party:Republican Party (United States)

The 1880 United States presidential election in Missouri took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for president and vice president.

Missouri voted for the Democratic nominee, Winfield Scott Hancock, over the Republican nominee, James A. Garfield by a margin of 13.84%.

See also