a)   nZn= 13/65 = 0,2 (mol)

Zn   + 2HCl → ZnCl2 +H2

0,2  →0,4

 mHCl= 0,4. 36,5=14,6(gam)

b)   nH2=0,4/2  =0,2 (mol)

mZn= 0,2 .65=13 (g)

mHCl  = 0,4.36,5 =14,6(gam)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

nH₂=6.72/22,4=0,3(mol)
PTHH: 2Al +   6HCl --> 2AlCl₃ + 3H₂
bài ra:  0,2 <-- 0,6  <--   0,2   <-- 0,3  /mol
a) mHCl = 0,6.36,5=21,9(g)
b) mAl = 0.2.24 = 4,8(g)
 

\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)

              0,2<--0,6<----------0,2<------0,3    (mol)

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)

\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

PTHH :

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,4        1,2         0, 4          0,6

\(m_{Al}=0,4.27=10,8\left(g\right)\)

\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

\(a)n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2      0,4             0,2         0,2

\(m_{FeCl_2}=0,2.127=25,4g\\ b.m_{HCl}=0,4.36,5=14,6g\)

5,Hg :)?

\(2Al+6HCl->2AlCl_3+3H_2\\ n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\cdot\dfrac{7,437}{24,79}=0,2mol\)