1868_United_States_presidential_election_in_Rhode_Island

1868 United States presidential election in Rhode Island

1868 United States presidential election in Rhode Island

Election in Rhode Island

The 1868 United States presidential election in Rhode Island took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Main article: 1868 United States presidential election
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Rhode Island voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 32.98%.

With 66.49% of the popular vote, Rhode Island would be Grant's fifth strongest victory in terms of popular vote percentage after Vermont, Massachusetts, Kansas and Tennessee.[1]

Results

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See also

References

  1. [1]
    "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. [2]
    "1868 Presidential General Election Results - Rhode Island".
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