Đáp án:

Bài 11:

`n_{Na_2CO_3}=m/M=(10,6)/(106)= 0,1 (mol)`

`456ml = 0,456l`

`=> C_{M_{Na_2CO_3}}=n/V=(0,1)/(0,456)= 0,22 (M)`

Bài 12:

`m_{\text{dd HCl}}=m_{ct} + m_{dm}=75 + 225= 300 (g)`

`=> \text{C%}_{HCl}=(m_{ct})/(m_{dd}) xx 100\text{%}=(75)/(300) xx 100\text{%}= 25\text{%}`

Bài 13:

`m_{FeCl_2}=(m_{dd} xx \text{C%})/(100\text{%})=(200 xx 9,525\text{%})/(100\text{%})= 19,05 (g)`

`=> n_{FeCl_2}=m/M=(19,05)/(127)= 0,15 (mol)`

`#Ly`

Bài 11:

`n_(Na_2CO_3` `=` `(10,6)/(106)=0,1(mol)`

Đổi $456ml=0,456l$

`C_MNa_2CO_3` `=` `(0,1)/(0,456)≈0,22(M)`

Bài 12:

`m_(ddB` `=` `75+225=300(g)`

`C%_(HCl` `=` `(75)/(300).100%=25%`

Bài 13:

`m_(FeCl_2` `=` `(200.9,525)/(100)=19,05(g)`

`n_(FeCl_2` `=` `(19,05)/(127)=0,15(mol)`