\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(\rightarrow nNaOH=\dfrac{8}{40}=0,2mol\rightarrow nH_2SO_4=0.1mol\)

\(\rightarrow VdungdichH_2SO_4=\dfrac{0,1}{1}=0,1l\)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{H_+}=\dfrac{1}{2}n_{OH^-}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

\(\Rightarrow V=0,1\left(l\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{0,5}=0,4(l)\)

Bảo toàn nguyên tố Cl, Fe:

\(n_{HCl}=2n_{FeCl_2}=2n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V=0,4\left(l\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{HCl}=3n_{Al}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{0,5}=0,6(l)\)

Bảo toàn nguyên tố Cl, Al:

\(n_{HCl}=n_{Cl}=3n_{AlCl_3}=3n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

`1)`

`n_{Al}={2,7}/{27}=0,1(mol)`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,1->0,15->0,05->0,15(mol)`

`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`

`->V=150`

`V'=V_{H_2}=0,15.22,4=3,36(l)`

`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`

`2)`

`n_{Fe}={2,8}/{56}=0,05(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,05->0,1->0,05->0,05(mol)`

`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`

`->V=100`

`V_{H_2}=0,05.22,4=1,12(l)`

`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`

Đề câu 2 sao khỏi làm đi

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{Fe}=0,1\left(mol\right);n_{HCl}=0,1.2=0,2\left(mol\right)\\ a,V_{ddHCl}=\dfrac{0,2}{4}=0,05\left(l\right)=50\left(ml\right)\\ b,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Bài 2. Cho 8g Fe2O3 tác dụng vừa đủ với dd HCl 20% (D = 1,1g/ml). Hãy tính: a. Thể tích dd HCl đã dùng b. Nồng độ % dd thu được sau phản ứng

a) \(n_{Fe_2O_3}=0,05\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=6n_{Fe_2O_3}=0,3\left(mol\right)\)

=> \(m_{ddHCl}=\dfrac{0,3.36,5}{20\%}=54,75\left(g\right)\)

=> \(V_{HCl}=\dfrac{m}{D}=\dfrac{54,75}{1,1}=49,77\left(g\right)\)

b) \(m_{ddsaupu}=8+54,75=62,75\left(g\right)\)

\(C\%_{FeCl_3}=\dfrac{0,05.2.162,5}{62,75}.100=25,9\%\)

a: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1       0,2        0,1            0,1

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

=>\(n_{H_2}=0.1\left(mol\right)\)

=>\(n_{HCl}=0.2\left(mol\right)\)

\(V_{HCl}=\dfrac{0.2}{4}=0.05\left(lít\right)\)

b: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)

a) $Mg + 2HCl \to MgCl_2 + H_2$
$n_{MgCl_2} = \dfrac{4,75}{95} = 0,05(mol)$

$n_{HCl} = 2n_{MgCl_2} = 0,1(mol)$
$m_{dd\ HCl} = \dfrac{0,1.36,5}{14,6\%} = 25(gam)$
$\Rightarrow V_{dd\ HCl} = \dfrac{25}{1,12} = 22,32(ml)$

b) $n_{Mg} = n_{H_2} = n_{MgCl_2} = 0,05(mol)$
$\Rightarrow m_{dd\ sau\ pư} = 0,05.24 + 25 - 0,05.2  = 26,1(gam)$

$C\%_{HCl} = \dfrac{4,75}{26,1}.100\% = 18,2\%$