Question

hòa tan 75ml dd NaOH 2M vào dd FeCl3 1m

a) Viết PTHH

b) Tính nồng độ mol sau PƯ

* đề ko cho ml của FeCl3

Answer 1

a)

\(n_{NaOH}=0,075.2=0,15\left(mol\right)\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\)

0,15------>0,05----------------------->0,15

b)

\(V_{dd.FeCl_3}=\dfrac{0,05}{1}=0,05\left(l\right)\\ V_{dd.sau.pư}=0,05+0,075=0,125\\ C_{M\left(NaCl\right)}=\dfrac{0,15}{0,125}=1,2M\)

Answer 2

\(V_{NaOH}=75ml=\dfrac{75}{1000}\left(l\right)=0,075\left(l\right)\)

=> \(n_{NaOH}=CM.V=2.0,075=0,15\left(mol\right)\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

0,15         0,05             0,05            0,15  (mol)

=> \(V_{FeCl_3}=\dfrac{n}{CM}=\dfrac{0,05}{1}=0,05\left(l\right)\)

\(V_{NaCl}=0,075+0,05=0,125\left(l\right)\)

\(CM_{NaCl}=\dfrac{n}{V}=\dfrac{0,15}{0,125}=1,2M\)