a) PTHH: Zn + H2SO4 -> ZnSO4 + H2

nH2= 0,15(mol)

=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)

b) mZn=0,15.65=9,75(g)

c) CMddH2SO4= 0,15/ 0,05=3(M)

d) mZnSO4= 161. 0,15=24,15(g)

\(PTHH:Zn+2HCl->ZnCl_2+H_2\)

ap dung DLBTKL ta co

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)

a) m HCl \(0,5.\left(36,5\right)=18,25g\)

b) Zn +2HCl -> ZnCl2 +H2

  0,25...0,5...................0,25

V H2 = 0,25.22,4 =5,6 lít 

m Zn = 0,25.65=16,25g 

a) Có: \(n_{HCl}=0,5mol\)

\(\Rightarrow m_{HCl}=0,5.\left(1+35,5\right)=18,25g\)

b)

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

Có: \(n_{HCl}=0,5mol\)

Theo PTHH:

\(\Rightarrow n_{H_2}=0,25mol\)

\(\Rightarrow V_{H_2}=0,25.22,4=5,6l\)

     \(n_{Zn}=0,25mol\)

\(\Rightarrow m_{Zn}=0,25.65=16,25g\)

nZn = 52 : 65 = 0,8 (mol) 
pthh : Zn + 2HCl ---> ZnCl2 +H2
         0,8--->1,6----------------->0,8 (mol) 
=> mHCl = 1,6 .  36,5 = 58,4 (g) 
VH2 = 0,8 . 22,4 = 17,92 (l) 
nFe3O4 = 9,28 : 232 = 0,04 (mol ) 
pthh : Fe3O4 + 4H2 -t--> 3Fe + 4H2O 
LTL : 
0,04/1  <   0,8/4  => H2 DU
theo pthh , nFe = 3nFe3O4 = 0,12 (mol) 
=> m Fe = 0,12 . 56= 6,72 (g)

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

 Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)

b, m_Zn = 0,15.65 = 9,75 (g)

c, C_{M (H2SO4)} = 0,15/0,05 = 3 M

d, m_ZnSO4 = 0,15.161 = 24,15 (g)

Bạn tham khảo nhé!

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

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

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

         0,2--->0,2--------->0,2------>0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)

`Zn+H_2SO_4->ZnSO_4+H_2`(to)

0,45-------------------0,45------0,45mol

`n_(Zn)=(29,25)/65=0,45mol`

`m_(ZnSO_4)=0,45.161=72,45g`

`V_(H_2)=0,45.22,4=10,08l`

c) `H_2+CuO->Cu+H_2O`(to)

       0,45--------0,45 mol

`n_(Cu)=40/80=0,5 mol`

=>Cu dư , 0,05 mol

`m_(chất rắn)=0,45.64+0,05.80=32,8g`

\(n_{Zn}=\dfrac{m}{M}=\dfrac{29,25}{65}=0,45\left(mol\right)\)

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

                    1           1               1           1

                   0,45     0,45           0,45       0,45

b) \(m_{ZnSO_4}=n.M=0,45.\left(65+32+16.4\right)=51,03\left(g\right)\\ V_{H_2}=n.24,79=0,45.24,79=11,1555\left(l\right)\) 

c) \(n_{CuO}=\dfrac{m}{M}=\dfrac{40}{\left(64+16\right)}=0,5\left(mol\right)\)

 \(PTHH:CuO+H_2\rightarrow Cu+H_2O\) 

                   1         1          1        1

                  0,5     0,5        0,5    0,5

\(m_{Cu}=0,5.64=32\left(g\right).\)