`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?

a) 2Al + 6HCl -----> 2AlCl3 + 3H2
Theo pt: 2 : 6 : 2 : 3 (mol)
Theo đb: 0,4 : 1,2 (mol)
Số mol H2 là:
nH2 = \(\frac{0,4\cdot3}{2}=0,6\left(mol\right)\)
Thể tích khí H2 (đktc) là:
VH₂ = n.22,4 = 0,6 . 22,4 = 13,44 (lít)
b) Số mol HCl là:
nHCl = \(\frac{0,4\cdot6}{2}=1,2\left(mol\right)\)
Khối lượng chất tan HCl là:
mct HCl = n.M = 1,2 . 36.5 = 43,8 (g)
Khối lượng dd HCl là:
mdd HCl = \(\frac{43,8\cdot100\%}{10,95\%}=400\left(g\right)\)
Khối lượng chất tan AlCl3 là:
mct AlCl3 = n.M = 0,4.133,5 = 53,4 (g)
Khối lượng dd AlCl3 là:
mdd AlCl3 = mAl + mHCl - mH₂ = 10,8 + 400 - 1,2 = 409,6 (g)
Nồng độ phần trăm dd AlCl3 là:
C% = \(\frac{53,4}{409,6}\cdot100\%\approx13,04\%\)

nAl= 10.8/27=0.4 mol

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

0.4___1.2_____0.4_____0.6

VH2= 0.6*22.4=13.44l

mHCl= 1.2*36.5=43.8g

mddHCl= 43.8*100/10.95=400g

mH2= 0.6*2=1.2g

mdd sau phản ứng= 10.8+400-1.2=409.6g

mAlCl3= 0.4*133.5=53.4g

C%AlCl3= 53.4/409.6*100%= 13.03%

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1........2\)

\(0.1......0.4\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)

\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)

\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)

\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)

d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)

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

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

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

a. 13,44 l

b. 13,35 %

Đây là hóa mà

mình ko hiểu sao ra đc 13,35 %