Fe+2HCl->FeCl2+H2

0,125---0,25--0,125----0,125---

n Fe=11.2\56=0,2 mol

n HCl=0,25.1=0,25 mol

=> lập tỉ lệ : 0,2\1>0,25\2

=>HCl hết

=>VH2=0,125.22,4=2,8l

=>m Fe=0,125.56=7g

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{HCl}=CM.V_{dd}=1.0,25=0,25\left(mol\right)\)

PTHH:\(2Fe+6HCl\rightarrow2FeCl_3+3H_2\)

TPƯ:   0,2       0,25

PƯ:     0,08    0,25         0,08         0,125

SPƯ:   0,12        0            0,08       0,125

\(V_{H_2}=n.22,4=0,125.22,4=2,8\left(l\right)\)

\(m_{Fedư}=n.M=0,12.56=6,72\left(g\right)\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư

\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,5}{2}>\dfrac{0,1}{1}\Rightarrow Zn.hết,HCldư\\ n_{HCl\left(dư\right)}=0,5-2.0,1=0,3\left(mol\right)\\ m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ n_{HCl}=0,2.2,5=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,2}{1}>\dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl,pư}=0,2.2=0,4mol\\ m_{HCl,dư}=\left(0,5-0,4\right).36,5=3,65g\)

nFe = 2.8/56 = 0.05 (mol) 

nHCl = 14.6/36.5 = 0.4 (mol) 

Fe + 2HCl => FeCl2 + H2 

1.........2

0.05......0.4 

LTL : 0.05/1 < 0.4/2 

=> HCl dư 

mHCl (dư) = ( 0.4 - 0.1 ) * 36.5 = 10.95 (g)

VH2 = 0.05*22.4 = 1.12 (l)

nHCl (dư) = 0.4 - 0.1 = 0.3 (mol)  

mFe cần thêm = 0.3/2 * 56 = 8.4 (g) 

a)

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

b)

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

Ta thấy : \(\dfrac{n_{Zn}}{1} = 0,2 > \dfrac{n_{HCl}}{2} = 0,15\) nên Zn dư.

Theo PTHH :

\(n_{Zn\ pư} = 0,5n_{HCl} = 0,15(mol)\\ \Rightarrow n_{Zn\ dư} = 0,2 - 0,15 = 0,05(mol)\\ \Rightarrow m_{Zn\ dư} = 0,05.65 = 3,25(gam)\)

c)

Ta có :

\(n_{H_2} = n_{Zn\ pư} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\)

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)

Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl}=4.100:1000=0,4\left(mol\right)\)

a. PTHH: MgO + 2HCl ---> MgCl₂ + H₂O

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)

Vậy HCl dư.

=> \(n_{dư}=\dfrac{0,1.2}{0,4}=0,5\left(mol\right)\)

=> \(m_{dư}=0,5.36,5=18,2\left(g\right)\) 

b. Ta có: \(V_{dd_{MgCl_2}}=V_{HCl}=\dfrac{100}{1000}=0,1\left(lít\right)\)

Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)

=> \(C_{M_{MgCl_2}}=\dfrac{0,1}{0,1}=1M\)