Gọi số mol Fe, Cu là a, b (mol)

=> 56a + 64b = 29,6 (1)

Có: 56a - 64b = 4 (2)

(1)(2) => a = 0,3 (mol); b = 0,2 (mol)

PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

                           0,4<------0,3

            CuO + H₂ --t^o--> Cu + H₂O

                      0,2<------0,2

=> V_H2 = (0,4 + 0,2).22,4 = 13,44 (l)

a) 4,45(g) kim loại + O₂ --> 6,05(g) oxit

b) 

Theo ĐLBTKL: m_{Kim loại} + m_O2 = m_oxit 

=> 4,45 + m_O2 = 6,05

=> m_O2 = 1,6 (g)

1)

Xét tỉ lệ: \(\dfrac{3}{2}< \dfrac{2}{1}\) => Hiệu suất tính theo SO₂

\(n_{SO_2\left(pư\right)}=\dfrac{3.75}{100}=2,25\left(mol\right)\)

PTHH: 2SO₂  + O₂ -----V₂O₅,t^o-----> 2SO₃

            2,25--------------------->2,25

=> n_SO3 = 2,25 (mol)

2) Gọi số mol SO₂ pư là a (mol)

PTHH: 2SO₂  + O₂ -----V₂O₅,t^o-----> 2SO₃

Trc pư:    3          2                                   0

Pư:          a-->0,5a-------------------->a

Sau pư: (3-a)   (2-0,5a)                           a

=> (3-a) + (2-0,5a) + a = 4,25

=> a = 1,5 (mol)

=> \(H=\dfrac{1,5}{3}.100\%=50\%\)

1) 

a) Gọi số mol Mg, Al, Zn là a, b,c (mol)

=> \(\left\{{}\begin{matrix}24a+27b+65c=19,46\left(1\right)\\24a=27b\left(2\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{16,352}{22,4}=0,73\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

              a--->2a--------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

              b-->3b-------------->1,5b

            Zn + 2HCl --> ZnCl₂ + H₂

             c---->2c------------>c

=> a + 1,5b + c = 0,73 (3)

(1)(2)(3) => a = 0,27 (mol); b = 0,24 (mol); c = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,27.24}{19,46}.100\%=33,3\%\\\%m_{Al}=\dfrac{0,24.27}{19,46}.100\%=33,3\%\\\%m_{Zn}=\dfrac{0,1.65}{19,46}.100\%=33,4\%\end{matrix}\right.\)

b) n_HCl = 2a + 3b + 2c = 1,46 (mol)

=> \(V_{dd.HCl}=\dfrac{1,46}{2}=0,73\left(l\right)\)

a, \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

0,5--->1------------->0,5

Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                   0,5----->0,5

b, \(V_{O_2}=1.22,4=22,4\left(l\right)\)

c, \(m_{CaCO_3}=0,5.100=50\left(g\right)\)

a, \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH

            0,2------------------>0,4

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{12,4+50}.100\%=25,64\%\)

b, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂

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

\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40+16}{100+16+4,6-0,1.2}.100\%==20\%\)

c, \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)

PTHH: 

2Na + 2HCl ---> 2NaCl + H₂

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

2Na + 2H₂O ---> 2NaOH + H₂

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

\(\Rightarrow m_{dd}=9,2+100-\left(0,1+0,1\right).2=108,8\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,2.58,5}{108,8}.100\%=10,75\%\\C\%_{NaOH}=\dfrac{0,2.40}{108,8}.100\%=7,35\%\end{matrix}\right.\)

NaCl - muối - natri clorua

CaSO₄ - muối - canxi sunfat

MgCl₂ - muối - magie clorua

ZnSO₄ - muối - kẽm sunfat

Cả 4 chất trên đều là muối có tên gọi theo thứ tự lần lượt là  natri clorua, canxi sunfat, magie clorua và kẽm sunfat.

a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2A + 2nHCl --> 2AlCl_n + nH₂

           \(\dfrac{0,5}{n}\)<-------------------0,25

=> \(M_A=\dfrac{14}{\dfrac{0,5}{n}}=28n\left(g/mol\right)\)

Xét n = 2 thỏa mãn => M_A = 56 (g/mol)

=> A là Fe

b) 

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,25<-0,5<----0,25<---0,25

=> n_{HCl(thực tế)} = \(\dfrac{0,5.110}{100}=0,55\left(mol\right)\)

=> m_{HCl(thực tế)} = 0,55.36,5 = 20,075 (g)

=> \(m_{dd.HCl}=\dfrac{20,075.100}{18,25}=110\left(g\right)\)

c) V_dd = \(\dfrac{110}{1,2}=\dfrac{275}{3}\left(ml\right)=\dfrac{11}{120}\left(l\right)\)

\(C_{M\left(dd.HCl.bđ\right)}=\dfrac{0,55}{\dfrac{11}{120}}=6M\)

- dd sau pư chứa HCl dư và FeCl₂

\(C_{M\left(FeCl_2\right)}=\dfrac{0,25}{\dfrac{11}{120}}=\dfrac{30}{11}M\)

\(C_{M\left(HCl.dư\right)}=\dfrac{0,55-0,5}{\dfrac{11}{120}}=\dfrac{6}{11}M\)

\(a,\left(1\right)4Na+O_2\underrightarrow{t^o}2Na_2O\\ \left(2\right)Na_2O+H_2O\rightarrow2NaOH\\ \left(3\right)2Na+2H_2O\rightarrow2NaOH+H_2\)

\(b,\left(1\right)4P+5O_2\underrightarrow{t^o}2P_2O_5\\ \left(2\right)P_2O_5+3H_2O\rightarrow2H_3PO_4\\ c,\left(1\right)Mg+2HCl\rightarrow MgCl_2+H_2\\ \left(2\right)2H_2+O_2\underrightarrow{t^o}2H_2O\\ \left(3\right)H_2O+SO_3\rightarrow H_2SO_4\\ \left(4\right)H_2SO_4+Zn\rightarrow ZnSO_4+H_2\\ \left(5\right)H_2+FeO\underrightarrow{t^o}Fe+H_2O\)

\(d,\left(1\right)2H_2O\underrightarrow{dp}2H_2+O_2\\ \left(2\right)H_2+FeO\underrightarrow{t^o}Fe+H_2O\\ \left(3\right)3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ e,\left(1\right)2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \left(2\right)2Ca+O_2\underrightarrow{t^o}2CaO\\ \left(3\right)CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

a.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)

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

0,2                                     0,2      ( mol )

\(V_{H_2}=0,2.24,79=4,958l\)

b.\(n_{Na}=\dfrac{6,9}{23}=0,3mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,3                                         0,15   ( mol )

\(V_{H_2}=0,15.24,79=3,7185l\)

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

  0,05                                              0,075      ( mol )

\(V_{H_2}=0,075.24,79=1,85925l\)