nNaOH = 0.4; nAlCl3 = 0.12

AlCl3 + 3NaOH ---> Al(OH)3+ 3NaCl

0.12----0.36 --------0.12 ----- 0.36

Al(OH)3 + NaOH ---> NaAlO2 + 2H2O

0.04 <----0.04 -------0.04

nA(OH)3 sau = 1.17/78 = 0.015 < 0.04

=> 2 TH mddY = 200 + 150 + m - 0.08*78 - 1.17 = m + 342.59

TH1: HCl thiếu

HCl + NaAlO2 + H2O ---> Al(OH)3 + NaCl

0.015--0.015

C%NaCl = 0.375*58.5/345.59 = 6.35%

TH2: HCl dư

HCl + NaAlO2 + H2O ---> Al(OH)3 + NaCl

0.04--0.04 --------------> 0.04-----0.04

sau pứ này Al(OH)3 bi htan 1 phần = 0.04 - 0.015 = 0.025 mol Al(OH)3 + 3HCl

---> AlCl3 + 3H2O

0.025 ----0.075 mddY = 0.115*36.5/0.1825 + 342.59 = 365.59g

Tổng nNaCl = 0.4 --> C% = 0.4*58.5/365.59 = 6.40%

=> Đáp án B

Chất nao sau đây la etylamin? A. C2H7N.B. C2H3NH2.C. CH3NH2