\(nCuO=\dfrac{8}{80}=0,1\left(mol\right)\)

\(nH_2SO_4=\dfrac{19,6}{98}=0,2\left(mol\right)\)

\(LTL:\dfrac{0,1}{1}< \dfrac{0,2}{1}\)

=> CuO pứ đủ , H2SO4 dư

CuO+H₂SO₄ -> CuSO₄+H₂O

0,1      0,1          0,1         0,1

\(nH_2SO_{4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(mH_2SO_{4\left(dư\right)}=0,1.98=9,8\left(g\right)\)

c1:

\(m\left(muối\right)=mCuSO_4=0,1.160=16\left(g\right)\)

c2:

\(mH_2O=0,1.18=1,8\left(g\right)\)

BTKL:

mCuO+mH₂SO₄   = m CuSO₄+ mH₂O

8          +  (19,6-9,8)  = m CuSO4  + 1, 8

=> mCuSO4 = 8 + ( 19,6 - 9,8 ) - 1,8 = 16 (g)

\(n_{Al_2O_3}=\dfrac{51.1000}{102}=500\left(mol\right)\)

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

500------------------->500

\(m_{Al_2\left(SO_4\right)_3}=1500.342=171000\left(g\right)=171\left(kg\right)\)

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

nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g 

\(n_{Al_2O_3}=\dfrac{20.4}{102}=0.2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{17.64}{98}=0.18\left(mol\right)\)

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

\(1.............3\)

\(0.2...........0.18\)

\(LTL:\dfrac{0.2}{1}>\dfrac{0.18}{3}\Rightarrow Al_2O_3dư\)

\(m_{Al_2\left(SO_4\right)}_{_3}=0.06\cdot342=20.52\left(g\right)\)

Gọi số mol của Al₂O₃, Fe₂O₃ là 2a, 3a (mol)

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

              2a---->6a

            Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

               3a----->9a

=> 6a + 9a = 0,3

=> a = 0,02 

=> \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,04\left(mol\right)\\n_{Fe_2O_3}=0,06\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,04.102}{0,04.102+0,06.160}.100\%=29,825\%\\\%m_{Fe_2O_3}=\dfrac{0,06.160}{0,04.102+0,06.160}.100\%=70,175\%\end{matrix}\right.\)

\(n_{Al2O3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Pt : \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

          1              3                    1                3

        0,2           0,6                   0,2

\(n_{H2SO4}=\dfrac{0,2.3}{1}=0,6\left(mol\right)\)

⇒ \(m_{H2SO4}=0,6.98=58,8\left(g\right)\)

\(n_{Al2\left(SO4\right)3}=\dfrac{0,6.1}{3}=0,2\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,2.342=68,4\left(g\right)\)

 Chúc bạn học tốt

\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ a,n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{3}{4}.0,3=0,225\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,225.22,4=5,04\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{1}{4}.0,3=0,075\left(mol\right)\\ n_{H_2SO_4}=3.0,075=0,225\left(mol\right)\\ m_{H_2SO_4}=m=0,225.98=22,05\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{74,37}{24,79}=3\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=2\left(mol\right)\)

\(\Rightarrow m_{Al}=2.27=54\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=3.98=294\left(g\right)\)