\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{13}{65}=0,2mol\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ c)n_{HCl}=0,2.2=0,4mol\\ C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ d)m_{ZnCl_2}=0,2.136=27,2g\)

1////          nAl=0,4mol   

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

0,4mol      1,2mol         0,4mol    0,6 mol

a/ V_H2=0,6.22,4=13,44 l

b/ V=1,2/2=0,6 l

C_AlCl3=0,4/0,6=2/3 M

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

nAl= 0,04(mol)

PTHH: 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

0,04___________0,06___0,02_____0,06(mol)

a) V(H2, đktc)=0,06.22,4=1,344(l)

b) VddH2SO4= 0,06/2=0,03(l)=30(ml)

c) VddAl2(SO4)3=VddH2SO4=0,03(l)

=>CMddAl2(SO4)3=0,02/0,03=2/3(M)

\(n_{Al}=\dfrac{1.08}{27}=0.04\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0.04......0.06.............0.02...........0.06\)

\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.06}{2}=0.03\left(l\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.02}{0.03}=\dfrac{2}{3}\left(M\right)\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH :

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,25      0,5                   0,25

\(a,m_{Fe}=0,25.56=14\left(g\right)\)

\(b,C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂↑

b. Ta có: \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{200:1000}=1,5M\)

=> \(n_{H_2SO_4}=0,3\left(mol\right)\)

Ta lại có: \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\)

Vậy H₂SO₄ dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

=> \(V_{H_2}=0,25.22,4=5,6\left(lít\right)\)

c. Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,25\left(mol\right)\)

=> \(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)

d. Ta có: \(V_{dd_{ZnSO_4}}=0,2\left(lít\right)\)

=> \(C_{M_{ZnSO_4}}=\dfrac{0,25}{0,2}=1,25M\)

a) 2NaOH + Cl₂ --> NaCl + NaClO + H₂O

b) \(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 2NaOH + Cl₂ --> NaCl + NaClO + H₂O

              0,1<----0,05--->0,05----->0,05

=> \(V_{ddNaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\)

c) 

\(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,05}{0,1}=0,5M\\C_{M\left(NaClO\right)}=\dfrac{0,05}{0,1}=0,5M\end{matrix}\right.\)