\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{13}{65}=0,2mol\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ c)n_{HCl}=0,2.2=0,4mol\\ C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ d)m_{ZnCl_2}=0,2.136=27,2g\)

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

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

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M) 

n Al=0,25 mol

2Al + 6HCl → 2AlCl₃ + 3H₂

0,25---0,75------0,25------0,375 mol

=>VH2=0,375.22,4=8,4l

=>m AlCl3=0,25.133,5=33,375g

=>CM HCl=\(\dfrac{0,75}{2}\)=0,375M

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

a. PTHH: Na₂SO₃ + 2HCl ---> 2NaCl + SO₂ + H₂O

Theo PT: \(n_{SO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)

=> \(V_{SO_2}=0,1.22,4=2,24\left(lít\right)\)

b. Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,1=0,2\left(mol\right)\)

=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

=> \(C_{\%_{HCl}}=\dfrac{7,3}{150}.100\%=4,87\%\)

c. Ta có: \(m_{dd_{NaCl}}=n_{Na_2SO_{3_{PỨ}}}=50\left(g\right)\)

Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)

=> \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

=> \(C_{\%_{NaCl}}=\dfrac{11,7}{50}.100\%=23,4\%\)

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

câu 1 
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,25   0,5         0,25       0,25 
\(m_{FeCl_2}=0,25.127=31,75g\\ V_{H_2}=0,25.22,4=5,6\\ C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5M\)
câu 2 
1 ) \(m_{\text{dd}}=35+100=135g\\ 2,C\%=\dfrac{204}{204+100}.100=60\%\\ =>m\text{dd}=\dfrac{100.204}{60}=340g\)
 

\(1,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

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

           0,3---->0,6------------------>0,3

\(2,C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\\ 3,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)

V H2 = 0,2.22,4 = 4,48(lít)

b)

n HCl = 2n Fe = 0,2.2 = 0,4(mol)

=> CM HCl = 0,4/0,4 = 1M

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư

Theo PTHH :

n CuO pư = n Cu = n H2 = 0,2(mol)

n Cu dư = 0,8 - 0,2 = 0,6(mol)

Vậy : 

%m Cu = 0,2.64/(0,2.64 + 0,6.80)   .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%