a)\(PTHH:Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)

b)\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(m\right)\);\(n_{H_2SO_4}=\dfrac{1,57}{98}=0,16\left(m\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

ta có tỉ lệ:\(\dfrac{0,3}{1}>\dfrac{0,16}{1}->Zndư\)

\(n_{Zn\left(dư\right)}=0,3-0,16=0,14\left(m\right)\)

\(m_{Zn\left(dư\right)}=0,14.65=9,1\left(g\right)\)

c)\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ        :1       1                1              1

số mol   :0,16   0,16           0,16         0,16

\(V_{H_2}=0,16.22,4=3,584\left(l\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)

\(PT:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(\dfrac{n_{Zn\left(ĐB\right)}}{n_{Zn\left(PT\right)}}=\dfrac{0,03}{1}>\dfrac{n_{H_2SO_4\left(ĐB\right)}}{n_{H_2SO_4}\left(PT\right)}=\dfrac{0,016}{1}\)

\(\Rightarrow\) Zn dư , H₂SO₄ hết , tính theo H₂SO₄  

b, Theo PT : \(n_{zn}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow m_{Zn\left(pứ\right)}=n\cdot M=0,016\cdot32=0,512\left(g\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=m_{Zn\left(ĐB\right)}-n_{Zn\left(Pứ\right)}=1,95-0,512=1,438\left(g\right)\)

c, Theo PT : \(n_{H_2}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow V_{H_{2\left(đktc\right)}}=n\cdot22,4=0,016\cdot22,4=0,3584\left(l\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo bài ra ta có :

\(m_{Zn}=1,95\left(g\right)\Rightarrow n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(m_{H_2SO_4}=1,57\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{m_{H_2SO_4}}{M_{H_2SO_4}}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)

Ta có tỉ lệ số mol chia cho hệ số :

\(\dfrac{0,03}{1}>\dfrac{0,016}{1}\)

\(\Rightarrow\) H2SO4 phản ứng hết / Zn phản ứng dư

\(\Rightarrow\) Tính theo số mol của H2SO4

⇒ \(n_{Zn}\) tham gia phản ứng là : \(n_{Zn}\approx0,016\left(mol\right)\Rightarrow m_{Zn}\approx0,016.32\approx0,512\left(g\right)\Rightarrow m_{Zn}\left(dư\right)=1,438\left(g\right)\)

\(V_{H_2}=0,016.22,4\approx0,3584\left(l\right)\)

\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ m_{H_2SO_4}=\dfrac{22,05.20}{100}=4,41\left(g\right)\\ n_{H_2SO_4}=\dfrac{4,41}{98}=0,045\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) 
\(LTL:\dfrac{0,03}{1}< \dfrac{0,045}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,03\left(mol\right)\\ m_{H_2SO_4\left(d\right)}=\left(0,045-0,03\right).98=1,47\left(g\right)\\ m_{\text{dd}}=1,95+22,05-\left(0,03.2\right)=23,94\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,03.136}{23,94}.100\%=17\%\)

\(a,n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ n_{H_2SO_4}=\dfrac{22,05}{98}=0,225\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

bđ       0,03   0,225

pư       0,03    0,03 

spư     0         0,195       0,03        0,03

\(b,m_{H_2SO_4\left(dư\right)}=0,195.98=19,11\left(g\right)\\ c,m_{dd}=1,95+22,05-0,03.2=23,94\left(g\right)\\ C\%_{ZnSO_4}=\dfrac{0,03.161}{23,94}.100\%=20,18\%\)

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

   \(m_{H_2SO_4}=147.10\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,1       0,1            0,1       0,1

Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ Zn hết, H₂SO₄ dư

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

c) m_{dd sau pứ} = 6,5 + 147 - 0,1.2 = 153,3 (g)

\(C\%_{ddZnSO_4}=\dfrac{0,1.161.100\%}{153,3}=10,502\%\)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,15-0,1\right).98.100\%}{153,3}=3,196\%\) 

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(n_{Fe_3O_4}=\dfrac{18,56}{232}=0,08\left(mol\right)\)

PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Xét tỉ lệ: \(\dfrac{0,08}{1}>\dfrac{0,2}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ        1     :     1       :       1            :  1

n(mol)     0,25-->0,25------->0,25------>0,25

\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ m_{ZnSO_4}=n\cdot M=0,25\cdot\left(65+32+16\cdot4\right)=40,25\left(g\right)\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\), ta được Zn dư.

Theo PT: \(n_{Zn\left(pư\right)}=n_{H_2SO_4}=0,015\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,03-0,015=0,015\left(mol\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=0,015.65=0,975\left(g\right)\)

c, \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,015.22,4=0,336\left(l\right)\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

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

   0,05    0,1          0,05       0,05

b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)

c)\(V_{H_2}=0,05\cdot22,4=1,12l\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)

Phản ứng thế

\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)

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

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