a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

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

0,1                         0,1      0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

                  0,15              0,1              ( mol )

\(m_{Fe}=0,1.56=5,6g\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\a, PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\\ c,C_{MddH_2SO_4}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\Tacó: n_{Fe}=n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\ \Rightarrow C\%_{H_@SO_4}=\dfrac{0,2.98}{200}.100=9,8\%\)

2Al + 3H2SO4 -----> Al2(SO4)3 + 3H2

nAl = 7,1/27 = 71/270 ( mol)

=> nH2 = 71/180 ( mol)

=> VH2= 8,86 lit

=> m muối=71\540 .342=44,967g

\(n_{Al}=\dfrac{7,1}{27}=\dfrac{71}{270}\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
            \(\dfrac{71}{270}\)                      \(\dfrac{71}{540}\)          \(\dfrac{71}{180}\) 
\(V_{H_2}=\dfrac{71}{540}.22,4=3l\\ m_{Al_2\left(SO_4\right)_3}=342.\dfrac{71}{180}=134,9g\)

a2Al+3H2SO4(dung dịch pha loãng)->Al2(SO4)3+3H2↑

b, nAl= 7,1/27=0,2 mol

Theo PTHH ta có: nH2=3/2 nAl= 3/2 . 0,2 = 0,3 mol

=> VH2(dktc)= 0,3 . 22,4= 6,72 lít

c, nAl2SO4= 1/2 . nAl= 1/2 . 0,2=0,1 mol

=> mAl2SO4=nAL2SO4+ MAl2SO4= 0,1 .  150 =15 g

a) \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

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

Xét tỉ lệ: \(0,2>\dfrac{0,3}{2}\Rightarrow\) Fe dư

Theo PTHH: \(n_{Fe\left(p\text{ư}\right)}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe\left(d\text{ư}\right)}=\left(0,2-0,15\right).56=2,8\left(g\right)\)

c) \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

Fe+H2SO4->FeSO4+H2

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

n Fe=0,1 mol

CMH2SO4=\(\dfrac{0,1}{0,1}=1M\)

=>VH2=0,1.22,4=2,24l

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)

nZn = 6.5/65 = 0.1 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.1.......0.2...................0.1

VddHCl = 0.2/2 = 0.1 (l) 

nFe = 3/56 (mol) 

Fe2O3 + 3H2 -to-> 2Fe + 3H2O

.................9/112........3/56

H% = 9/112 / 0.1 * 100% = 80.35%

a) Zn + 2HCl $\to$ ZnCl2 + H2

b) n Zn = 6,5/65 = 0,1(mol)

Theo PTHH : n HCl = 2n Zn = 0,2(mol)

=> V dd HCl = 0,2/2 = 0,1(lít)

c) n Fe = 3/56 (mol)

Fe2O3 + 3H2 $\xrightarrow{t^o}$ 2Fe + 3H2O

Theo PTHH :

n H2 = 3/2 n Fe = 9/112(mol)

Vậy :

H = $\dfrac{ \dfrac{9}{112} }{0,1}$ .100% = 80,36%

nZn=0,1(mol)

a) PTHH: Zn + 2 HCl -> ZnCl2 + H2

0,1_________0,2______0,1____0,1(mol)

b) VddHCl=0,2/2=0,1(l)

c) PTHH: 3 H2 + Fe2O3 -to-> 2 Fe + 3 H2O

nFe(LT)= 2/3. 0,1= 1/15(mol)

nFe(TT)= 3/56(mol)

=> H=[(3/56):(1/15)].100=80,357%

a. \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

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

=> \(C\%_{H_2SO_4}=\dfrac{0,3.98.100}{200}=14,7\%\)