1873 Rhode Island gubernatorial election
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The 1873 Rhode Island gubernatorial election was held on April 2, 1873, in order to elect the governor of Rhode Island. Republican nominee Henry Howard defeated Democratic nominee Benjamin G. Chace.[1]
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April 2, 1873
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 County results Howard: 60â70% 70â80% >90% | |||||||||||||||||
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General election
On election day, April 2, 1873, Republican nominee Henry Howard won the election by a margin of 5,870 votes against his opponent Democratic nominee Benjamin G. Chace, thereby retaining Republican control over the office of governor. Howard was sworn in as the 32nd governor of Rhode Island on May 6, 1873.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Henry Howard | 9,656 | 71.68 | |
| Democratic | Benjamin G. Chace | 3,786 | 28.11 | |
| Scattering | 29 | 0.21 | ||
| Total votes | 13,471 | 100.00 | ||
| Republican hold | ||||