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

PTHH:

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

0,2                      0,2          0,3 

\(V_{H_2}=n.22,4=6,72\left(l\right)\)

\(m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)

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

b) \(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\) ;    \(n_{HCl}=\dfrac{m}{M}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

 \(2Al\)    \(+\)   \(6HCl\)    →    \(2AlCl_3\)  \(+\)  \(3H_2\)

Tỉ lệ:   \(\dfrac{0,2}{2}>\dfrac{0,5}{6}\)                                          ⇒  Al dư, tính theo HCl

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

           \(0,5\)    →   \(\dfrac{1}{6}\)    →   \(0,25\)       ( mol )

\(V_{H_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)

c) \(m_{AlCl_3}=n.M=\dfrac{1}{6}.\left(27+35,5.3\right)=22,25\left(g\right)\)

2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g

B nha

a. 2Al + 6HCl -> 2AlCl₃ + 3H₂

b. n_Al = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)

\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)

c. \(n_{HCl}=3n_{Al}=3.0,3=0,9\left(mol\right)=>m_{HCl}=0,9.36,5=32,85\left(g\right)\)

Vậy m = 32,85

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

n HCl = o,2 mol 

        2 Al +6 HCl →2AlCl₃ + 3H2

bđ:   0,1

đang bận !

nAl = \(\frac{4,05}{27}=0,15mol\)

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

0,15 0,45 0,15 0,225 (mol)

a) nHCl = 0,45 mol

=> mHCl = 0,45 . 36,5 = 16,425 g

b) nAlCl3 = 0,15 mol

=> mAlCl3 = 0,15 . 133,5 = 20,025 g

c) nH2 = 0,225 mol

=> mH2 = 0,225 . 2 = 0,45 g

=> V_H2 = 0,225 . 22,4 = 5,04 lit

Ta có: \(n_{Al}=\dfrac{4,86}{27}=0,18\left(mol\right)\)

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

a, \(n_{HCl}=3n_{Al}=0,54\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,54}{0,15}=3,6\left(M\right)\)

b, \(n_{AlCl_3}=n_{Al}=0,18\left(mol\right)\Rightarrow m_{AlCl_3}=0,18.133,5=24,03\left(g\right)\)

c, \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,27\left(mol\right)\Rightarrow V_{H_2}=0,27.24,79=6,6933\left(l\right)\)

mai mình thi r giúp mình vs

\(a.2Al+6HCl->2AlCl_3+3H_2\)

        0,1.....................................0,1..........0,15

b. n Al = \(\frac{2,7}{27}=0,1mol\)

m AlCl3 = 0,1(27+35,5.3)=13,35 g

V H2 = 0,15.22,4=3,36 lít

cảm ơn bạn nha

\(\begin{array}{l} n_{H_2}=\dfrac{6,72}{22,4}=0,3\ (mol)\\ PTHH:\\ 2Al+6HCl\to 2AlCl_3+3H_2\uparrow\ (1)\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\ (2)\\ Theo\ pt\ (1):\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\ (mol)\\ \Rightarrow m_{Al}=0,2\times 27=5,4\ (g).\\ \Rightarrow m_{Al_2O_3}=15,6-5,4=10,2\ (g)\\ \Rightarrow n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\ (mol)\\ \Rightarrow \sum n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,2+2\times 0,1=0,4\ (mol)\\ \Rightarrow m_{AlCl_3}=0,4\times 133,5=53,4\ (g)\end{array}\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)

b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)

\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)

0,1    0,05

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)