... H 4 dư :(a – b) mol C H O −+ )( 7 3 ba ) 9 ( b b + ) 9 ( b b + 11 11 11 3 44 12 baa == 18 2b b 9 56b 63 16 63 56 : 9 : 12 ⋅ bbb // CO 2 15 ,0 2,0 H 2 O 15 ,0 2.25,0 Ni t 0 C Ni t 0 C C 2 H 2 pư ... 8,0 5,3 610 0 3, 2 91 00 = K 2 CO 3 mol9,0 13 8 2 ,12 4 = H 2 SO 4 mol x x 25,0 98 10 0 5,2 410 0 = HCl pư AgNO 3 CO 2 K 2 CO 3 pư H 2 SO 4 K 2 CO 3 ddH 2 SO 4 CO 2 H 2 O AgCl ở cốcA 1/ 2dd A HNO 3 (1/ 2dd A) ... (3,4): n = 1/ 2n + 1/ 2n = 1/ 2.0,3 +1/ 2.0 ,1= 0,2 < 0,65. Vâỵ: K 2 CO 3 dư, ta có: n = n = 0,2 mol (0,25 điểm) m = 213 ,2 + 63,55 – ( 0,2x 44) = 267 ,95 gam (0,25 điểm) m = 213 ,2 – 63,55= 1 49, 65 gam....