\(n_{HCl}=0,05.1,5=0,075\left(mol\right);n_{H_2}=0,03\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Vì:\dfrac{0,075}{6}>\dfrac{0,03}{3}\Rightarrow HCldư\\ a,n_{H_2\left(TT\right)}=\dfrac{0,075}{2}=0,0375\left(mol\right)\\ H=\dfrac{0,03}{0,0375}.100\%=80\%\\ b,n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,03=0,02\left(mol\right)\Rightarrow m_{Al}=0,02.27=0,54\left(g\right)\\ c,m_{AlCl_3}=0,02.133,5=2,67\left(g\right)\)

a) nAl=2,7/27=0,1(mol)

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

0,1_________0,3___0,1_____0,15(mol)

b) mHCl=0,3.36,5=10,95(g)

c) mAlCl3=0,1.133,5=13,35(g)

d) V(H2,đktc)=0,15.22,4=3,36(l)

a. PTHH: 2Al + 6HCl \(\rightarrow\) 2AlCl\(_3\) + 3H\(_2\)

b. \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1\) (mol)

\(n_{HCl}=\dfrac{0,1.6}{2}=0,3\) (mol)

\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\) (g)

c. \(n_{AlCl_3}=\dfrac{0,1.2}{2}=0,1\) (mol)

\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,1.133,5=13,35\) (g)

d. \(n_{H_2}=\dfrac{0,1.2}{3}=0,15\) (mol)

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36\) (l)

Ta có: \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

a, Theo PT: \(n_{H_2\left(LT\right)}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2\left(LT\right)}=0,15.24,79=3,7185\left(l\right)\)

\(\Rightarrow H=\dfrac{V_{H_2\left(TT\right)}}{V_{H_2\left(LT\right)}}.100\%=\dfrac{2,479}{3,7185}.100\%\approx66,67\%\)

b, \(n_{HCl}=\dfrac{6}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n_{HCl}}{C_{M_{HCl}}}=\dfrac{0,3}{1,5}=0,2\left(l\right)\)

c, \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

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

\(0.1........0.3............0.1......0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

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

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

\(n_{HCl}=\dfrac{25,55}{35,6}=0,7\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,7}{3}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,7-0,6=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

a) nHCl=14,6/36,5=0,4(mol); nAl=2,7/27=0,1(mol)

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

Ta có: 0,1/2 < 0,4/6

=>Al hết, HCl dư, tính theo nAl

nHCl(dư)= 0,4- 0,1.6/2=0,1(mol)

=>mHCl(dư)=0,1.36,5=3,65(g)

b) nAlCl3=nAl=0,1(mol)

=> mAlCl3=133,5.0,1=13,35(g)

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

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

Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(LT\right)}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\\n_{HCl\left(LT\right)}=2n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\)

Mà: H% = 80%

\(\Rightarrow\left\{{}\begin{matrix}n_{Al\left(TT\right)}=\dfrac{0,2}{80\%}=0,25\left(mol\right)\\n_{HCl\left(TT\right)}=\dfrac{0,6}{80\%}=0,75\left(mol\right)\end{matrix}\right.\)

⇒ m_Al = 0,25.27 = 6,75 (g)

m_HCl = 0,75.36,5 = 27,375 (g)