a) Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

                0,1---->0,3------->0,1

=> m = 0,1.342 = 34,2 (g)

c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

0,1------0,3------0,1-------------------0,3

n Al2O3=0,1 mol

m Al2(SO4)3=0,1.342=34,2g

C%=\(\dfrac{0,3.98}{120}100=24,5\%\)

\(a.Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2 O\\ b.n_{Al_2O_3}=0,1\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ c.n_{H_2SO_4}=3n_{Al_2O_3}=0,3\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100=24,5\%\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.6\cdot0.1=0.06\left(mol\right)\)

\(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)

\(2............3\)

\(0.1.........0.06\)

\(LTL:\dfrac{0.1}{2}>\dfrac{0.06}{3}\Rightarrow Aldư\)

\(m_{Al\left(dư\right)}=\left(0.1-0.04\right)\cdot27=1.62\left(g\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.02}{0.1}=0.2\left(M\right)\)

a) $2Al + 3CuSO_4 \to Al_2(SO_4)_3 + 3Cu$

b) n CuSO4 = 0,1.0,6 = 0,06(mol)

Theo PTHH : 

n Al pư = 2/3 n CuSO4 = 0,04(mol)

m Al dư = 2,7 - 0,04.27 = 1,62(gam)

c)

n Al2(SO4)3 = 1/2 n Al = 0,02(mol)

CM Al2(SO4)3 = 0,02/0,1 = 0,2M

a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

             0,4--->0,6-------------------->0,6

=> V_H2 = 0,6.22,4 = 13,44 (l)

c) \(V_{dd.H_2SO_4}=\dfrac{0,6}{1}=0,6\left(l\right)\)

d) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\) => Fe₂O₃ hết, H₂ dư

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

               0,1----------------->0,2

=> m_Fe = 0,2.56 = 11,2 (g)

\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1mol\\ 2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\\ 0,1................0,15.............0,05............0,3\\ C_{\%H_2SO_4}=\dfrac{0,15.98}{300}\cdot100\%=4,9\%\\ C_{\%Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{7,8+300}\cdot100\%=5,56\%\)

n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)

n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)

       \(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)

bđ    0,1............0,5

pư    0,1............0,3..................0,1

spu   0 ................0,2................0,1 

=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O

m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)

m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)

m dd = \(490+10,2=500,2g\)

% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)

% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)