$8)$Dung dịch axit làm quỳ tím chuyển sang màu đỏ. 

$9)$ `n_(FeO)=(7,2)/72=0,1(mol)`

$PT:FeO+H_2\xrightarrow{t^0}Fe+H_2O$

Theo `PT,n_(FeO)=n_(H_2)=0,1(mol)`

`=>V_(H_2)(đktc)=0,1.22,4=2,24(l)`

$10)$`n_(CuO)=48/80=0,6(mol)`

$PT:CuO+H_2\xrightarrow{400^oC}Cu+H_2O$

Theo `PT,n_(CuO)=n_(H_2)=0,6(mol)`

`=>V_(H_2)(đktc) =0,6.22,4=13,44(l)`

$11)$

`m_(dd)=5+45=50(g)`

`m_(ct)=m_(NaCl)=5(g)`

`=>C%NaCl=5/50. 100%=10%`

$12)$`n_(CaCl_2)=(2,22)/111=0,02(mol)`

Đổi $100(ml)=0,1(l)$

`=>CM_(CaCl_2)=(0,02)/(0,1)=0,2(M)`

Đáp án+Giải thích các bước giải:

$8.$ Dung dịch Axit làm quỳ tím chuyển sang màu đỏ.

$9.$ $n_{FeO}=\dfrac{m_{FeO}}{M_{FeO}}=$ $\dfrac{7,2}{72}=0,1$ $mol$

PTHH: $FeO+H_2\xrightarrow{t^o}Fe+H_2O$

$n_{H_2}=n_{FeO}=0,1$ $mol$

$\Rightarrow$ $V_{H_2}=n_{H_2}\times22,4=$ $0,1\times22,4=2,24l$

$10.$ $n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=$ $\dfrac{48}{80}=0,6$ $mol$

PTHH: $CuO+H_2\xrightarrow{400^oC}Cu+H_2O$

$n_{H_2}=n_{CuO}=0,6$ $mol$

$\Rightarrow$ $V_{H_2}=n_{H_2}\times22,4=$ $0,6\times22,4=$ $13,44l$

$11.$ $m_{dd_{NaCl}}=5+45=50g$

$C\text{%}_{NaCl}=\dfrac{m_{NaCl}\times100\text{%}}{m_{dd_{NaCl}}}=$ $\dfrac{5\times100\text{%}}{50}=10\text{%}$

$12.$ $n_{CaCl_2}=\dfrac{m_{CaCl_2}}{M_{CaCl_2}}=$ $\dfrac{2,22}{111}=$ $0,02$ $mol$

$V=100ml=0,1l$

$C_{M_{CaCl_2}}=\dfrac{n_{CaCl_2}}{V_{CaCl_2}}=$ $\dfrac{0,02}{0,1}=0,2M$

$^{_{_o}}tintinday$