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cho 6,5 g Zn tác dụng với dung dịch chứa 12g HCl. Thể tích khí H2 ( ở đktc ) là

Sagot :

nhinguyenthao62

Answer: nHCl = 12/36,5= 0,38(mol)

nZn = 6,5/65 = 0,1(mol)

Ta có pt: Zn + 2HCl --> ZnCl2 + H2

Vì 2n Zn = 0,2 < n HCl = 0,38 nên HCl dư

Theo PTHH :

n H2 = n Zn = 0,1(mol)

V H2 = 0,1.22,4 = 2,24(lít)

Explanation:

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