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

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

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

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

2Al  +  6HCl   →  2AlCl₃  +  3H₂

nAl = \(\dfrac{3,375}{27}\)= 0,125 mol

a) Theo tỉ lệ phản ứng => nH₂ = \(\dfrac{3}{2}\)nAl = 0,1875 mol

<=> V H₂ = 0,1875.22,4 = 4,2 lít

b) nAlCl₃ = nAl = 0,125 mol

=> mAlCl₃ = 0,125 . 133,5 = 16,6875 gam

Zn+2HCl->ZnCl2+H2

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

nZn=0,2 mol

=>m Hcl=0,4.36,5=14,6g

m muối=0,2.136=27,2g

=>VH2=0,2.22,4=4,48l

`Zn + 2HCl -> ZnCl_2 + H_2` `\uparrow`

`n_(Zn) = 13/65 = 0,2 mol`.

`n_(HCl) = 0,4 mol`.

`m_(HCl) = 0,4 xx 36,5 = 14,6g`.

c, `m_(ZnCl_2) = 0,2 xx 127 = 25,4 g`.

`d, V_(H_2) = 0,2 xx 22,4 = 4,48l`.

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ a,m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2..............0,4.............0,2...............0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{MgCl_2}=95.0,2=19\left(g\right)\\ c,a=C_{MddHCl}=\dfrac{0,4}{0,2}=2\left(M\right)\)

n_Al = 5.4 / 27 = 0.2 (mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

0.2......0.6............0.2.......0.3

a) V_H2 = 0.3 * 22.4 = 6.72 (l) 

b) m_AlCl3 = 0.2 * 133.5 = 26.7 (g) 

c) VddHCl = 0.6 / 1.5 = 0.4 (l) 

d) C_MAlCl3 = 0.2 / 0.4 = 0.5 (M) 

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)

\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2       0,3                  0,1                  0,3

\(V_{H_2}=0,3.24,79=7,437l\\ b/C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c/m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(5.a/n_{MgO}=\dfrac{4}{40}=0,1mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)

PTPỨ: Zn + 2HCl ---> ZnCl2 + H2
nZn = 6,5/65 = 0,1 (mol)
Theo ptpứ: nHCl = 2nZn = 0,2 (mol)

=> mHCl = 0,2 x 36,5 = 7,3 (g)
nH2 = nZn = 0,1 (mol)
=> VH2(đktc) = 0,1 x 22,4 = 2,24 (l)
mdd HCl đã dùng = 7,3 x 7,3% = 0,5329(g)

Câu c bạn viết rõ ra đi ak. CM hay C% và của chất j

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

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

\(0.2.........0.6...........0.2\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

\(V_{dd_{HCl}}=\dfrac{0.6}{3}=0.2\left(l\right)\)