\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

           0,1 <----------------------------- 0,1

=> m_Zn = 0,1.65 = 6,5 (g)

Zn+H2SO4->ZnSO4+H2

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

n H2=0,1 mol

=>m Zn=0,1.65=6,5g

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ pthh:Zn+H_2SO_4->ZnCl_2+H_2\)
         0,1                                        0,1 
=>\(m_{Zn}=0,1.65=6,5 \left(g\right)\)

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

Phương trình hóa học phản ứng

Zn + H₂SO₄ ---> ZnSO₄ + H₂

1   :      1       :       1         :  1

0.2                     0,2          0,2

mol                    mol          mol

=> \(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

b) \(m_{ZnSO_4}=n.M=0,2.161=32,2\left(g\right)\)

c) Ta có \(C\%=\frac{m_{ct}}{m_{dd}}.100\%=24,5\%\)

=> \(m_{ct}=\frac{C\%.m_{dd}}{100\%}=\frac{24,5\%.200}{100\%}=49\left(g\right)=m_{H_2SO_4}\)

=> \(m_{H_2O}=151\left(g\right)\)

=> \(n_{H_2SO_4}=\frac{m}{M}=\frac{49}{98}=0,5\)(mol) 

Dễ thấy \(\frac{n_{Zn}}{1}< \frac{n_{H_2SO_4}}{1}\)

=> H₂SO₄ dư 0,5 - 0,2 = 0,3 (mol) 

=> \(m_{H_2SO_4\text{ dư }}=n.M=0,3.98=29,4\left(g\right)\); \(m_{H_2SO4\text{ tham gia}}=n.M=0,2.98=19,6\)(g)

Áp dụng đinhk luật bảo toàn khối lượng 

=> \(m_{H_2SO_4}+m_{Zn}=m_{ZnSO4}+m_{H_2}\)

=> \(m_{H_2}=m_{H_2SO_4}+m_{Zn}-m_{ZnSO_4}=19,6+13-32,2=0,4\left(g\right)\)

=> \(m_{saupư}=m_{ZnSO_4}+m_{H_2SO_4\text{ dư}}+m_{H_2O}-m_{H_2}=32,2+29,4+151-0,4=232,2\left(g\right)\)

=> \(C\%_{H_2SO_4}=\frac{m_{ct}}{m_{sau\text{ pư}}}.100\%=\frac{29,4}{232,2}.100\%=12,66\%\)

\(C\%_{ZnSO_4}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{32,2}{232,2}.100\%=13,87\%\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,1     0,2         0,1           0,1 
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\\ m_{HCl}=\dfrac{\left(0,2.36,5\right).100}{24,5}=29,795\left(g\right)\\ m_{\text{dd}}=5,6+29,795-\left(0,1.2\right)=35,195\left(g\right)\\ C\%=\dfrac{0,1.127}{35,195}.100\%=36\%\)

4.1)

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

PTHH: Zn + H₂SO_{4 (loãng)} ---> ZnSO₄ + H₂ 

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

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

c) \(m_{ddH_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)

4.2)

Gọi kim loại hóa trị II là R; \(n_{AgCl}=\dfrac{2,87}{143,5}=0,02\left(mol\right)\)
PTHH: \(RCl_2+2AgNO_3\rightarrow R\left(NO_3\right)_2+2AgCl\downarrow\)

           0,01<-----------------------------------0,02

=> \(M_{RCl_2}=\dfrac{0,95}{0,01}=95\left(g/mol\right)\)

=> \(M_R=95-71=24\left(g/mol\right)\)

Mà R có hóa trị II => R là Magie (Mg)

Câu 5:

a) Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Al_2O_3}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)

=> 80a + 102b = 14,2 (1)

n_HCl = 0,2.3,5 = 0,7 (mol)

PTHH:
CuO + 2HCl ---> CuCl₂ + H₂O

a------>2a

Al₂O₃ + 6HCl ---> 2AlCl₃ + 3H₂O

b------->6b

b) 2a + 2b = 0,7 (2)
Từ (1), (2) => a = 0,05; b = 0,1

=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{14,2}.100\%=28,17\%\\\%m_{Al_2O_3}=100\%-28,17\%=71,83\%\end{matrix}\right.\)

a)

Chất rắn còn lại sau phản ứng là Cu vì Cu không phản ứng với dung dịch sunfuric 0,5M

\(Zn + H_2SO_4 \to ZnSO_4 + H_2\)

Theo PTHH : \(n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\)

\(\Rightarrow m_{Cu} = m_{hỗn\ hợp} - m_{Zn} = 10,5 - 0,1.65 = 4(gam)\)

b)

Ta có : \(n_{H_2SO_4} = n_{ZnSO_4} = n_{H_2} = 0,1(mol)\)

Suy ra :

\(V_{H_2SO_4} = \dfrac{0,1}{0,5} = 0,2(lít)\\ m_{ZnSO_4} = 0,1.161 = 16,1(gam)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
         0,1        0,2                        0,1 
\(V_{H_2}=0,1.22,4=2,24l\\ m_{HCl}=\left(0,2.36,5\right).10\%=0,73g\) 
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ LTL:\dfrac{0,1}{1}>\dfrac{0,1}{3}\) 
=> Fe2O3 dư 
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,067\left(mol\right)\\ m_{Fe}=0,067.56=3,73g\)

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

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

0,1      0,2                          0,1   ( mol )

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

\(m_{ddHCl}=\dfrac{0,2.36,5}{10\%}=73g\)

c.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

 0,1   >    0,1                                    ( mol )

                0,1                1/15                     ( mol )

\(m_{Fe}=\dfrac{1}{15}.56=3,73g\)

a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

b) Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)=n_{FeSO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,01\cdot152=1,52\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)

c) Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=0,01\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01\cdot98}{19,6\%}=5\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,9(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,9.98}{19,6\%}=450(g)\)