Số xấu quá em xem lại đề nha

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

a) \(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(n_{HCl}=\dfrac{m}{M}=\dfrac{18,5}{36,5}=\dfrac{37}{73}\left(mol\right)\)

Lập bảng: \(\dfrac{0,4}{1}>\dfrac{\dfrac{37}{73}}{2}\)

⇒ sau pư HCl hết, Fe dư

⇒ theo \(n_{HCl}\)

Theo PTHH: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{\dfrac{37}{73}}{2}=\dfrac{37}{146}\left(mol\right)\)

⇒ \(n_{Fe\left(dư\right)}=0,4-\dfrac{37}{146}=\dfrac{107}{730}\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=n.M=\dfrac{107}{730}.56=\dfrac{2996}{365}\left(g\right)\)

b) Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{\dfrac{37}{73}}{2}=\dfrac{37}{146}\left(mol\right)\)

\(\Rightarrow V_{H_2\left(đktc\right)}=n.22,4=\dfrac{37}{146}.22,4=\dfrac{2072}{365}\left(l\right)\)

c) Theo PTHH: \(n_{FeCl_2}=n_{H_2}=\dfrac{37}{146}\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=n.M=\dfrac{37}{146}.127=\dfrac{4699}{146}\left(g\right)\)

Fe+2HCl->Fecl₂+H₂

1--------0,2-----0,1----0,1

n Fe=\(\dfrac{5,6}{56}\)=0,1 mol

n HCl=\(\dfrac{36,5}{36,5}\)=1 mol

=>HCl dư :0,8mol

=>m _HCl=0,8.36,5=29,2g

=>m _FeCl2=0,1.127=12,7g

=>V_H2=0,1.22,4=2,24l

a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\); \(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1->0,2----->0,1--->0,1

=> \(m_{HCl\left(dư\right)}=\left(1-0,2\right).36,5=29,2\left(g\right)\)

b) \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

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

a) $Fe + 2HCl \to FeCl_2 + H_2$

b)

n Fe = 0,56/56 = 0,01(mol)

n HCl = 1,095/36,5=  0,03(mol)

Ta thấy :

n Fe / 1 = 0,01 < n HCl /2 = 0,015 => HCl dư

n HCl pư = 2n Fe = 0,02(mol)

=> m HCl dư = 1,095 - 0,02.36,5 = 0,365 gam

mFe= 8,4/56= 0,15 mol 

m HCl = 14,6/36,5=0,4 mol 

       PTHH:                Fe +2HCl →FeCl₂ +H₂

                       Bđ:    0,15   0,4             0     0          mol

                       Pứ:    o,15→0,3           0,15   0,15       mol

                       Sau pứ:0      0,1            0,15     0,15     mol 

a. HCl dư: m =0,1.36,5=3,65 g 

b. m FeCl₂ = 0,15.127=19,05 g 

c. m H2 = 0,15.2= 0,3 g 

V H2= 0,15.22,4=3,36 (l)

Bạn học lớp mấy rồi bạn ? aaaa

a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right);n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

Ta có: \(\dfrac{0,3}{1}< \dfrac{0,5}{1}\) ⇒ Fe hết, H₂SO₄ dư

PTHH:Fe + H₂SO₄ ----> FeSO₄ + H₂

Mol:    0,3      0,3                           0,3

\(m_{H_2SO_4dư}=\left(0,5-0,3\right).98=19,6\left(g\right)\)

b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

a. \(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{49}{98}=0,5\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

1    :     1                          :    1    (mol)

0,3  :   0,5  (mol)

-Chuyển thành tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,5}{1}\Rightarrow\)Fe phản ứng hết còn H₂SO₄ dư.

\(m_{H_2SO_4\left(lt\right)}=n.M=\dfrac{0,3.1}{1}.98=29,4\left(g\right)\)

\(m_{H_2SO_4\left(dư\right)}=m_{H_2SO_4\left(tt\right)}-m_{H_2SO_4\left(lt\right)}=49-29,4=19,6\left(g\right)\)

b. -Theo PTHH trên: \(n_{H_2\left(đktc\right)}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=n.M=0,3.22,4=6,72\left(l\right)\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right);n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\\ \Rightarrow Mgdư\\ \Rightarrow n_{Mg\left(p.ứ\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\\ b,m_{MgCl_2}=95.0,05=4,75\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

a) nMg=0,2(mol)

PTHH: Mg +2 HCl -> MgCl2 + H2

0,2________0,4____0,2_____0,2(mol)

b) V(H2,đktc)=0,2.22,4=4,48(l)

c) Zn + H2SO4 -> ZnSO4 + H2

0,2<------------------------------0,2(mol)

=>mZn=0,2.65=13(g)

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.2.................................0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

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

\(0.2......................................0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\)

Fe+2HCl->FeCl2+H2

0,6--------------------0,6

H2+CuO-to>Cu+H2O

0,6--------------0,6

n Fe=0,6 mol

=>VH2=0,6.22,4=13,44l

=>m Cu=0,6.64=38,4g

\(n_{Fe}=\dfrac{33,6}{56}=0,6mol\)

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

0,6       1,2          0,6       0,6

\(V_{H_2}=0,6\cdot22,4=13,44l\)

\(CuO+H_2\rightarrow Cu+H_2O\)

0,6       0,6      0,6

\(m_{Cu}=0,6\cdot64=38,4g\)