a) Số mol của Al là

\(n=\dfrac{m}{M}\) = \(\dfrac{4.68}{27}\)= 0.1733 mol

PTHH

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

            2 mol                    2 mol       3 mol

         0.1733 mol         0.1733 mol    0.26 mol

 Theo pthh \(n_{AlCl_3}\) = 0,1733 mol

vậy Khối lượng của AlCl₃ là

\(m=M\cdot n\) = 133.5 \(\cdot\) 0,1733 = 23.1355 mol

b) theo pthh \(n_{H_2}\) = 0.26 mol

vậy thể tích của H₂ ở đkc là 

\(V=n\cdot24.79\) = 0.26\(\cdot\)24.79 = 6.4454 (L)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b.n_{Al_2\left(SO_4\right)_3}=\dfrac{0,3}{3}=0,1mol\\ m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{6,5}{65}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1      0,2             0,1             0,1

\(m_{ZnCl_2}=0,1.136=13,6g\\ c)V_{H_2}=0,1.24,79=2,479l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100\%=3,65\%\)

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

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

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

______0,2_________________0,2 (mol)

b, V_H2 = 0,2.22,4 = 4,48 (l)

c, Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H₂ dư.

Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)

⇒ m_Fe = 0,1.56 = 5,6 (g)

Bạn tham khảo nhé!

a) Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

b) m_Zn = \(\dfrac{13}{65}\)=0,2 (mol)

         Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

(mol) 0,2 ----------------------> 0,2

\(V_{H_2}\)= 0,2 . 22,4 = 4,48(lít)

c)\(n_{FeO}\)=\(\dfrac{7,2}{72}\)=0,1 (mol)

         H₂ + FeO \(\underrightarrow{t^o}\)Fe + H₂O

(mol)         0,1----->0,1

m_Fe = 0,1 . 56 = 5,6(g)

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)

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

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

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

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

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

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

a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

b,\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Na_2CO_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c, \(n_{CO_2}=n_{Na_2CO_3}=0,2\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)