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) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{54}{27}=2\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{2.3}{2}=3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=3.22,4=67,2l\)

b) \(2H_2+O_2\rightarrow2H_2O\)

\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{30}{32}=0,94\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=\dfrac{0,94.2}{1}=1,88\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=1,88.18=33,84\left(g\right)\)

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ b)n_{O_2} =\dfrac{4,8}{32}=0,15(mol)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ \dfrac{n_{H_2}}{2} = 0,1 < \dfrac{n_{O_2}}{1} = 0,15 \to O_2\ dư\\ n_{H_2O} = n_{H_2} = 0,2(mol)\\ \Rightarrow m_{H_2O} = 0,2.18= 3,6(gam)\)

\(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,04   0,08                  0,04

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,08}{0,2}=0,4M\)

  \(V_{H_2}=0,04.22,4=0,896\left(l\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

\(\Rightarrow n_{HCl}=2n_{Zn}=2.\dfrac{2,6}{65}=0,08\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{0,08}{0,2}=0,4M\)

\(n_{H_2}=n_{Zn}=0,08\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,08.22,4=1,792\left(l\right)\)

a) Ta có: n_Zn= \(\frac{m_{Zn}}{M_{Zn}}=\frac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl -> ZnCl₂ + H₂

Theo PTHH và đề bài, ta có:

n_HCl = 2. n_Zn= 2.0,2=0,4 (mol)

Khối lượng HCl đã dùng:

m_HCl= n_HCl.M_HCl= 0,4. 36,5= 14,6 (g)

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

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Xét tỉ lệ:  \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

__________0,2-------------->0,1

=> V_H2 = 0,1.22,4 = 2,24(l)

b)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

______0,1->0,05

=> m_O2 = 0,05.22,4 = 1,12 (l)

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

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

           0,1-------------->0,1------>0,1

\(\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c) \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

     0,1<---0,1

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)

a: Zn+2HCl->ZnCl2+H2

0,2       0,4      0,2       0,2

mZnCl2=0,2*136=27,2(g)

b: V=0,2*22,4=4,48(lít)

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)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{H_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{1\cdot36,5}{20\%}=182,5\left(g\right)\\V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)