a.

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{29.2}{36.5}=0.8\left(mol\right)\)

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

Lập tỉ lệ : 

\(\dfrac{0.2}{2}< \dfrac{0.8}{6}\rightarrow HCldư\)

Khi đó : 

\(n_{AlCl_3}=0.2\left(mol\right),n_{H_2}=0.2\cdot\dfrac{3}{2}=0.3\left(mol\right)\)

\(n_{HCl\left(dư\right)}=0.8-0.2\cdot3=0.2\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.2\cdot36.5=7.3\left(g\right)\)

\(b.\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

\(c.\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

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

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

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

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

n HCl = o,2 mol 

        2 Al +6 HCl →2AlCl₃ + 3H2

bđ:   0,1

đang bận !

\(1.\\ a)M_x=\dfrac{1}{0,01}=100g/mol\\ b)n=\dfrac{1,2395}{24,79}=0,05mol\\ M=\dfrac{3,2}{0,05}=64\\ 2.\\ n_{Al}=\dfrac{2,7}{27}=0,1mol\\ n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\dfrac{0,1}{2}< \dfrac{0,4}{6}\Rightarrow HCl.dư\\ n_{H_2}=\dfrac{0,1.3}{2}=0,15mol\\ V_{H_2}=0,15.24,79=3,7185l\)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1         0,2          0,1          0,1

\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

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

b) Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow n_{HCl}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,3\cdot36,5}{100}\cdot100\%=10,95\%\)

c+d) Theo PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,3\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=104,8\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{26,7}{104,8}\cdot100\%\approx25,48\%\)