\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)

`n_[Al]=[2,7]/27=0,1(mol)`

`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`

`0,1`    `0,3`              `0,1`      `0,15`           `(mol)`

`a)V_[H_2]=0,15.22,4=3,36(l)`

`b)V_[dd HCl]=[0,3]/2=0,15(l)`

`=>C_[M_[AlCl_3]]=[0,1]/[0,15]~~0,67(M)`

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
          0,1        0,3            0,1            0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)

\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\) 

\(2Al+6HCl->2AlCl_3+3H_2\) (1)

theo (1) \(n_{H_2}=\frac{3}{2}n_{Al}=0,6\left(mol\right)\) 

=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(n_{Al}=\dfrac{4.5}{27}=\dfrac{1}{6}\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\dfrac{1}{6}.....0.5.......\dfrac{1}{6}.......0.25\)

\(m_{HCl}=0.5\cdot36.5=18.25\left(g\right)\)

\(m_{AlCl_3}=\dfrac{1}{6}\cdot133.5=22.25\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

Yêu cầu của đề là gì vậy em ?

nAl=1/6(mol)

PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

1/6________1/2(mol)

Ủa đề cứ thiếu thiếu , bắt tính gì làm gì em?

1////          nAl=0,4mol   

2Al     +      6HCl -----> 2AlCl3 + 3H2

0,4mol      1,2mol         0,4mol    0,6 mol

a/ V_H2=0,6.22,4=13,44 l

b/ V=1,2/2=0,6 l

C_AlCl3=0,4/0,6=2/3 M

n_Al=m/M=10,8/27=0,4 (mol)

PT:

2Al + 6HCl -> 2AlCl₃ + 3H₂\(\uparrow\)

2............6............2..............3 (mol)

0,4 -> 1,2 -> 0,4 ->0,6 (mol)

Khí thoát ra là H₂

V_H2=n.22,4= 0,6.22,4=13,44 (lít)

b) V_{d d HCl}=n.C_M=1,2.2=2,4 (lít)

C_{M AlCl3}=\(\dfrac{n}{V}=\dfrac{0,4}{2,4}\approx0,17\left(M\right)\)

nAl = 10,8/27 = 0,4 mol

PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2

Theo PTHH nH2 = 3/2nAl = 3/2 . 0,4 = 0,6 mol

=> VH2 = 0,6.22,4 = 13,44 lít

b:Theo PTHH nHCl = 3nAl = 0,4.3 = 1,2 mol

Vdd = n.CM = 1,2.2 = 2,4 lít

Theo PTHH nAlCl3 = nAl = 0,4 mol

CM của AlCl3 = 0,4/2,4 = 0,17 M

chúc bạn học tốt :))

câu 1 PTHH:Fe+2HCl\(\xrightarrow[]{}\)FeCl₂+H₂

n_Fe=\(\dfrac{5,6}{56}\)=0,1 mol

a) theo đầu bài ta có

n_Fe=n_H2=0,1 mol

lượng khí H2 tạo ra ở điều kiện tiêu chuẩn là

V=n.22,4

V_H2=0,1.22,4= 2,24 (l)

Cau 2

Ta có pthh

2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2

Theo đề bài ta có

nAl=\(\dfrac{10,8}{27}=0,4mol\)

a, Theo pthh

nH2=\(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,4=0,6mol\)

\(\Rightarrow\) VH2=0,6.22,4=13,44 l

b, Theo pthh

nHCl=\(\dfrac{6}{2}.nAl=\dfrac{6}{2}.0,4=1,2mol\)

\(\Rightarrow mHCl=1,2.36,5=43,8g\)

\(\Rightarrow m\text{dd}_{HCl}=\dfrac{mct.100\%}{C\%}=\dfrac{43,8.100\%}{10,95\%}=400g\)

Theo pthh

nAlCl3=nAl=0,4 mol

\(\Rightarrow\) mAlCl3=0,4.133,5=53,4 g

mdd\(_{AlCl3}\)= mAl + m\(_{\text{dd}HCl}\) - mck = 10,8 +400 - ( 0,6.2)=409,6 g

\(\Rightarrow\) Nồng độ % của chất sau phản ứng là :

C%=\(\dfrac{mct}{m\text{dd}}.100\%=\dfrac{53,4}{409,6}.100\%\approx13,04\%\)

Câu 3

Ta có pthh

2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2

Theo đề bài ta có

nAl=\(\dfrac{10,8}{27}=0,4mol\)

a, Theo pthh

nH2=\(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,4=0,6mol\)

\(\Rightarrow\) VH2=0,6.22,4=13,44 l

b, Theo pthh

nHCl=\(\dfrac{6}{2}nAl=\dfrac{6}{2}.0,4=1,2mol\)

\(\Rightarrow\) Vdd\(_{HCl}\)=\(\dfrac{nct}{CM}=\dfrac{1,2}{2}=0,6l=600ml\)

Theo pthh

nAlCl3=nAl=0,4 mol

\(\Rightarrow\) CM=\(\dfrac{n}{V}=\dfrac{0,4}{0,6}\approx0,67M\)

n_Al = 5.4 / 27 = 0.2 (mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

0.2......0.6............0.2.......0.3

a) V_H2 = 0.3 * 22.4 = 6.72 (l) 

b) m_AlCl3 = 0.2 * 133.5 = 26.7 (g) 

c) VddHCl = 0.6 / 1.5 = 0.4 (l) 

d) C_MAlCl3 = 0.2 / 0.4 = 0.5 (M) 

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)