14
12
11
10
Correct answer is A
K\(_{2}\)Cr\(_{2}\)O\(_{7}\) + HCl → Cl\(_{2}\) + KCl + CrCl\(_{3}\) + H\(_{2}\)O
Balancing the equation, we have
K\(_{2}\)Cr\(_{2}\)O\(_{7}\) + 14HCl → 3Cl\(_{2}\) + 2KCl + 2CrCl\(_{3}\) + 7H\(_{2}\)O
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