Romulo Rodrigues Mota vs Leandro Fabricio Campelo
5. Floripa Open 2019, 2019 · Result 1–0 · East Indian Defense (A48).
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Game details
- White
- Romulo Rodrigues Mota (1993)
- Black
- Leandro Fabricio Campelo (2201)
- Result
- 1–0
- Event
- 5. Floripa Open 2019
- Year
- 2019
- Opening
- East Indian Defense (A48)
About this chess game
This chess game between Romulo Rodrigues Mota (1993) and Leandro Fabricio Campelo (2201) was played at 5. Floripa Open 2019 in 2019 and finished 1–0. The opening was the East Indian Defense (A48). You can replay the full game move by move on the interactive board above, or open it on the CipherChess analysis board to study every move with the Stockfish engine.
Looking for more Romulo Rodrigues Mota games or Leandro Fabricio Campelo games? This Romulo Rodrigues Mota vs Leandro Fabricio Campelo encounter is one of millions of chess games indexed in the CipherChess mega database. Browse both players' full records, the openings they play most, and head-to-head results, then load any game onto the board to prepare your own lines against the East Indian Defense.
Frequently asked questions
Who won Romulo Rodrigues Mota vs Leandro Fabricio Campelo?
Romulo Rodrigues Mota vs Leandro Fabricio Campelo (2019) finished 1–0, a win for Romulo Rodrigues Mota.
What opening was played in Romulo Rodrigues Mota vs Leandro Fabricio Campelo?
The game opened with the East Indian Defense (ECO A48).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Romulo Rodrigues Mota vs Leandro Fabricio Campelo, or open it on the CipherChess analysis board to review it with the Stockfish engine.