a, Theo gt ta có: $n_{H_2}=0,15(mol);n_{O_2}=0,05(mol)$

$2H_2+O_2\rightarrow 2H_2O$

Sau phản ứng $H_2$ còn dư. Và dư 0,05.22,4=1,12(l)

b, Ta có: $n_{H_2O}=2.n_{O_2}=0,1(mol)\Rightarrow m_{H_2O}=1,8(g)$

nH2 = 3.36/22.4 = 0.15 (mol) 

nO2 = 1.12/22.4 = 0.05 (mol) 

2H2 + O2 -to-> 2H2O 

0.1___0.05_____0.1 

VH2 (dư) = ( 0.15 - 0.1) * 22.4 = 1.12 (l) 

mH2O = 0.1*18 = 1.8 (g) 

\(n_{KMnO_4}=\dfrac{18.96}{158}=0.12\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.12...........................................0.06\)

\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)

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

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(0.08.....0.06.......0.04\)

\(m_{Al\left(dư\right)}=\left(0.2-0.08\right)\cdot27=3.24\left(g\right)\)

\(m_{Al_2O_3}=0.04\cdot102=4.08\left(g\right)\)

2Cu+O2-to>2CuO

0,1-----0,05-----0,1

4P+5O2-to>2P2O5

n Cu=\(\dfrac{6,4}{64}\)=0,1 mol

=>VO2=0,05.22,4=1,12l

=>m CuO=0,1.80=8g

b) 

thiếu đề

a. \(n_{Cu}=\dfrac{6.4}{64}=0,1\left(mol\right)\)

PTHH : 2Cu + O₂ -> 2CuO

            0,1       0,05     0,1

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(m_{CuO}=0,1.81=8,1\left(g\right)\)

b. Thiếu số mol P 

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=\dfrac{245.20\%}{98}=0,5\left(mol\right)\)

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

0,2..........0,5

Lập tỉ lệ : \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\)

=> H2SO4 dư

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

=> V H2 = 0,3.22,4= 6,72(l)

\(m_{ddsaupu}=5,4+245-0,3.2=249,8\left(g\right)\)

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{249,8}.100=13,69\%\)

a) mH2SO4=20%.245=49(g) ->nH2SO4=49/98=0,5(mol)

nAl=5,4/27=0,2(mol)

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

Ta có: 0,2/2 < 0,5/3

=> H2SO4 dư, Al hết, tính theo nAl

=> nH2SO4(p.ứ)=nH2=3/2. nAl=3/2. 0,2= 0,3(mol)

=> nH2SO4(dư)=0,5 - 0,3=0,2(mol)

=>mH2SO4(dư)=0,2.98=19,6(g)

b) V(H2,đktc)=0,3.22,4=6,72(l)

c) nAl2(SO4)3= 1/2. nAl=1/2. 0,2=0,1(mol)

=>mAl2(SO4)3=342.0,1=34,2(g)

mddAl2(SO4)3=mAl+ mddH2SO4-mH2=5,4+245 - 0,3.2= 249,8(g)

=>C%ddAl2(SO4)3= (34,2/249,8).100=13,691%

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(0.1.....0.05\)

\(m_{O_2\left(dư\right)}=\left(0.5-0.05\right)\cdot32=14.4\left(g\right)\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

\(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{23,2}{160}=0,145mol\)

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

0,145    > 0,2                                  ( mol )

 1/15        0,2             2/15                 ( mol 0

Chất còn dư là \(Fe_2O_3\)

\(m_{Fe_2O_3\left(du\right)}=n_{Fe_2O_3\left(du\right)}.M_{Fe_2O_3}=\left(0,145-\dfrac{1}{15}\right).160=12,53g\)

\(m_{Fe}=n_{Fe}.M_{Fe}=\dfrac{2}{15}.56=7,4666g\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

LTL: \(\dfrac{1,5}{2}< 1,5\rightarrow O_2\) dư

Theo pt: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.1,5=0,75\left(mol\right)\\n_{H_2O}=n_{H_2}=1,5\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(1,5-0,75\right).32=24\left(g\right)\\V_{O_2}\left(1,5-0,75\right).22,4=16,8\left(l\right)\\m_{H_2O}=1,5.18=27\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(MOL\right)\) 
pthh: \(2H_2+O_2\underrightarrow{t^O}2H_2O\) 
LTL : \(\dfrac{1,5}{2}< \dfrac{1,5}{1}\) 
=> O2 dư , H2 hết 
theo pthh: nH2O = nH2 = 1,5 (mol) 
 => \(m_{H_2O}=1,5.18=27\left(g\right)\)

n_H = \(\frac{3,36}{22,4}=0,15\)(mol)

n_{O2 = \(\frac{3,36}{22,4}=0,15\)}

PTHH: 4H₂ + O₂ = 2H₂O

Tỉ lệ:     4         1        2

Theo PTHH và đề bài, ta có:

\(\frac{0,15}{4}< \frac{0,15}{1}\)

=> H₂ hết, O₂ dư

b) PTHH: 4H₂ + O₂ = 2H₂O

nH_2O = 0,15 : 4 x 2 = 0,075

mH₂O = 0,075 x 18 = 1,35 (g)