2CH3COOH+Fe->(CH3COO)2Fe+H2
0,4------------------0,2----0,2--------------0,2
n H2=\(\dfrac{4,958}{24,79}\)=0,2 mol
=>CM CH3COOH=\(\dfrac{0,4}{0,2}\)=2M
=>m (CH3COO)2Fe=0,2.174=34,8g
CH3COOH+NaOH->CH3COONa+H2O
0,1---------------0,1
n CH3COOH=2.0,05=0,1 mol
=>m NaOH=0,1.40=4g
=>m dd NAOH=400g
Đúng 1
Bình luận (2)