\(\left\{{}\begin{matrix}KBr\left(pư\right):x\left(mol\right)\\KBr\left(dư\right):y\left(mol\right)\end{matrix}\right.\)→ x + y = \(\dfrac{2,38}{119} = 0,02\)(1)
2KBr + Cl2 → 2KCl + Br2
x.......................x................(mol)
Suy ra : 74,5x + 119y = 1,7125(2)
Từ (1)(2) suy ra : x = 0,015 ; y = 0,005
Vậy : mKBr = 0,005.119 =0,595(gam).Đáp án B