nZn= 30/65= 6/13(mol)

PTHH: Zn +2 HCl -> ZnCl2 + H2

nHCl= 12/13(mol); nH2= 6/13(mol)

=> V(H2,đktc)=6/13 . 22,4=10,338(l)

mHCl=12/13 . 36,5=438/13(mol)

=> mddHCl= 438/13 : 15%=224,615(g)

Bn nào giúp mình với ạk

Bài 2

Bài 1

a/. Phương trình phản ứng hoá học: 
Fe + 2HCl --> FeCl2 + H2 
b/. nH2 = V/22,4 = 3,36/22,4 = 0,15 (mol) 
....... Fe.....+ 2HCl --> Fecl2 + H2 
TPT 1 mol....2 mol.................1 mol 
TDB x mol....y mol................0,15 mol 
nFe = x = (0,15x1)/1 = 0,15 (mol) 
mFe = n x M = 0,15 x 56 = 8,4 (g) 
c/. nHCl = y = (0,15x2)/1 = 0,3 (mol) 
CMHCl = n/V = 0,3/0,05 = 6 (M) 

\(n_{Al}=\frac{2,7}{27}=0,1mol\)

`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`

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

⇒ m_FeO = 12,6 - 5,4 = 7,2 (g)

c, Phần này đề cho dd NaOH dư hay vừa đủ bạn nhỉ?

d, Cho hh vào dd H₂SO₄ đặc nguội thì có khí thoát ra.

PT: \(2FeO+4H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=\dfrac{1}{2}n_{FeO}=0,05\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

 \(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,1      0,3         0,1       0,15

Ta có: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) ⇒ Al hết, HCl hết

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

c, m_{dd sau pứ} = 2,7 + 109,5 - 0,15.2 = 111,9 (g)

\(C\%_{ddAlCl_3}=\dfrac{13,35.100\%}{111,9}=11,93\%\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) \(\Rightarrow\) Al và HCl đều p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=111,9\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{111,9}\cdot100\%\approx11,93\%\)

a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)

c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

PTHH :

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,05         0,1         0,05

\(b,m_{CuO}=0,05.80=4\left(g\right)\)

\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)