Đáp án + Giải thích các bước giải:

Bài 1:

$a\big)Fe+2HCl\to FeCl_2+H_2$

$b\big)$

Theo PT: $n_{H_2}=n_{Fe}=\frac{5,6}{56}=0,1(mol)$

$\to V=V_{H_2(đktc)}=0,1.22,4=2,24(l)$

$c\big)$

Theo PT: $n_{FeCl_2}=0,1(mol);n_{HCl}=0,2(mol)$

$\to m_{FeCl_2}=0,1.127=12,7(g)$

$\to m_{HCl}=0,2.36,5=7,3(g)$

Bài 2:

$a\big)$

$2Na+2H_2O\to 2NaOH+H_2$

$2K+2H_2O\to 2KOH+H_2$

$b\big)$

$n_{Na}=\frac{4,6}{23}=0,2(mol)$

$n_K=\frac{3,9}{39}=0,1(mol)$

Theo PT: $\sum n_{H_2}=\frac{1}{2}n_{Na}+\frac{1}{2}n_K=0,15(mol)$

$\to V=V_{H_2(đktc)}=0,15.22,4=3,36(l)$

$c\big)$

Theo PT: $n_{NaOH}=0,2(mol);n_{KOH}=0,1(mol)$

$m_{dd\,spu}=4,6+3,9+100-0,15.2=107,9(g)$

$\to\begin{cases} C\%_{NaOH}=\dfrac{0,2.40}{107,9}.100\%≈7,41\%\\ C\%_{KOH}=\dfrac{0,1.56}{107,9}.100\%≈5,19\% \end{cases}$

Đáp án + Giải thích các bước giải:

 Bài 1: 

`a)` 

`Fe + 2HCl→ FeCl_2+ H_2`

`b)`

`n_{Fe} = {5,6}/{56} = 0,1(mol)`

`n_{H_2} = n_{Fe} = 0,1(mol)`

`V= 0,1 .22,4= 2,24(mol)`

`c)`

`n_{HCl} = 2n_{H_2} = 2 .0,1= 0,2(mol)`

`m_{HCl} = 0,2 .36,5= 7,3(g)`

`n_{FeCl_2} = n_{Fe} = 0,1(mol)`

`m_{FeCl_2} = 0,1 . 127= 12,7(g)`

Bài 2: 

`a)`

`2Na+ 2H_2O→ 2NaOH+ H_2` 

`2K+ 2H_2O→ 2KOH+ H_2`

`b)`

`n_{Na} = {4,6}/{23} = 0,2(mol)`

`n_{K} = {3,9}/{39} = 0,1(mol)`

`n_{H_2} = 1/2(n_{Na} + n_K)`

`= 1/2(0,2+ 0,1) = 0,15(mol)`

`V= 0,15 .22,4= 3,36(l)`

`c)`

`mdd= 4,6 + 3,9+ 100- 0,15 .2=108,2(g)`

`n_{NaOH} = n_{Na} = 0,2(mol)`

`n_{KOH} = n_K= 0,1(mol)`

`C%_{NaOH} = {0,2 .40}/{108,2} .100= 7,4%`

`C%_{KOH} = {0,1 . 56}/{108,2} . 100=5,2%`