Đáp án:

$a\big)V_{H_2(đktc)}=2,24(l)$

$b\big)C\%_{HCl}=7,3\%$

$c\big)C\%_{FeCl_2}\approx 12,05\%$

Giải thích các bước giải:

$a\big)$

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

$Fe+2HCl\to FeCl_2+H_2$

Theo PT: $n_{H_2}=n_{Fe}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

$b\big)$

Theo PT: $n_{HCl}=2n_{Fe}=0,2(mol)$

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

$\to C\%_{HCl}=\frac{7,3}{100}.100\%=7,3\%$

$c\big)$

Theo PT: $n_{FeCl_2}=n_{Fe}=0,1(mol)$

Ta có $m_{dd\,spu}=5,6+100-0,1.2=105,4(g)$

$\to C\%_{FeCl_2}=\frac{0,1.127}{105,4}.100\%\approx 12,05\%$

Đáp án:

$V_{H_2}$ = $2,24(lít)$

$\text{C% HCl = 7,3%}$

$\text{C% $FeCl_2$ = 12,04(%)}$

Giải thích các bước giải:

$\text{a,}$

$nFe$ = $\frac{5,6}{56}$ = $0,1(mol)$

$Fe + 2HCl → FeCl_2 + H_2$

$\text{$\scriptsize{0,1(mol) → 0,2(mol) → 0,1(mol) → 0,1(mol)}$}$

$\text{$V_{H_2}$ = 0,1 * 22,4 = 2,24(l)}$

$\text{b,}$

$\text{mHCl = 0,2 * 36,5 = 7,3(g)}$

$\text{C% HCl = $\frac{7,3}{100}$ * 100 = 7,3%}$

$\text{c,}$

$\text{m dung dịch sau phản ứng: mFe + mddHCl - m$H_2$ = 105,4(g)}$

$\text{m $FeCl_2$ = 0,1 * 127 = 12,7(g)}$

$\text{C% $FeCl_2$ = $\frac{12,7}{105,4}$ * 100 = 12,04(%)}$