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

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)

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

            \(FeO+H_2\xrightarrow[]{t^o}Fe+H_2O\) (2)

             \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\) (3)

             \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\) (4)

Theo PT (2),(3),(4): \(n_{O\left(pư\right)}=n_{H_2}=0,3\left(mol\right)\)

=> \(m_{giảm}=m_O=0,3.16=4,8\left(g\right)\)

n Mg=0,3 mol

Mg+2HCl->MgCl2+H2

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

->mgiảm=0,3.16=4,6g

Ta có: \(m_{giảm}=m_{O\left(oxit\right).tham.gia.phản.ứng}\)

=> \(n_O=\dfrac{8}{16}=0,5\left(mol\right)\)

PTHH:

\(Fe_3O_4+4H_2\xrightarrow[]{t^o}3Fe+4H_2O\\ ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)

Theo PT: \(n_{H_2}=n_{O\left(oxit\right)}=0,5\left(mol\right)\)

=> V_H2 = 0,5.22,4 = 11,2 (l)

`a) SO_3, SO_2, Al_2O_3`

`SO_3 + 2KOH -> K_2SO_4 + H_2O`

`SO_2 + 2KOH -> K_2SO_3 + H_2O`

`Al_2O_3 + 2KOH -> 2KAlO_2 + H_2O`

`b) K_2O, CuO, Al_2O_3,SO_3,Fe_2O_3`

`K_2O + H_2SO_4 -> K_2SO_4 + H_2O`

`SO_3 + H_2SO_4(đặc) -> H_2SO_4.nSO_3`

`Fe_2O_3 + 3H_2SO_4 -> Fe_2(SO_4)_3 + 3H_2O`

`Al_2O_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2O`

`c) SO_3, Al_2O_3`

`CuO + H_2SO_4 -> CuSO_4 + H_2O` nữa bạn nhé

a) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

b) \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)

Theo PTHH: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,4\left(mol\right)\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,4.400=160\left(g\right)\)

c) Theo PTHH: \(n_{H_2SO_4}=3.n_{Fe_2O_3}=1,2\left(mol\right)\)

=> \(m_{H_2SO_{\text{4}}}=1,2.98=117,6\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{117,6}{180}.100\%=65,33\%\)

a) \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

b) \(n_{H_2SO_4}=0,15.2=0,3\left(mol\right)\)

PTHH: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

               0,3<----0,3--------->0,3

=> m_ZnO = 0,3.81 = 24,3 (g)

c) \(m_{ZnSO_4}=0,3.161=48,3\left(g\right)\)

a)

\(n_{M_xO_y}=\dfrac{3,06}{x.M_M+16y}\left(mol\right)\)

PTHH: \(M_xO_y+2yHNO_3\rightarrow xM\left(NO_3\right)_{\dfrac{2y}{x}}+yH_2O\)

\(n_{M\left(NO_3\right)_{\dfrac{2y}{x}}}=\dfrac{5,22}{M_M+\dfrac{124y}{x}}\left(mol\right)\)

Theo PTHH: \(x.n_{M_xO_y}=n_{M\left(NO_3\right)_{\dfrac{2y}{x}}}\)

=> \(\dfrac{3,06x}{x.M_M+16y}=\dfrac{5,22}{M_M+\dfrac{124y}{x}}\)

=> \(3,06.x.M_M+379,44y=5,22.x.M_M+83,52y\)

=> \(M_M=68,5.\dfrac{2y}{x}\left(g/mol\right)\)

Xét \(\dfrac{2y}{x}=2\) thỏa mãn => M_M = 137 (g/mol)

=> M là Ba

b)

Chất rắn không tan sau khi hòa tan cặn vào dd HCl là tạp chất

=> \(\%m_{tạp.chất}=\dfrac{0,012}{7,05}.100\%=0,17\%\)

Ta có: \(m_{BaCO_3}=0,209-0,012=0,197\left(g\right)\Rightarrow n_{BaCO_3}=\dfrac{0,197}{197}=0,001\left(mol\right)\)

PTHH: \(BaO+CO_2\rightarrow BaCO_3\)

           0,001<--0,001<--0,001

=> \(\%m_{BaO\left(biến.thành.muối.cacbonat\right)}=\dfrac{0,001.153}{7,05-0,012}.100\%=2,174\%\)

\(m_{tăng}=m_{CO_2}+m_{H_2O}\)

=> \(0,001.44+18.n_{H_2O}=7,184-7,05\)

=> \(n_{H_2O}=0,005\left(mol\right)\)

PTHH: \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

          0,005<--0,005

=> \(\%m_{BaO\left(bị.hút.ẩm\right)}=\dfrac{0,005.153}{7,05-0,012}.100\%=10,87\%\)

anh bn à

Gọi \(\left\{{}\begin{matrix}n_{K_2CO_3}=a\left(mol\right)\\n_{KCl}=b\left(mol\right)\\n_{KHCO_3}=c\left(mol\right)\end{matrix}\right.\)

=> 138a + 74,5b + 100c = 39,09 (1)

PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)

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

             \(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)

                   c-------->c-------->c------>c

=> \(a+c=\dfrac{6,72}{22,4}=0,3\) (2)

dd sau pư chứa \(\left\{{}\begin{matrix}KCl:2a+b+c\left(mol\right)\\HCl_{dư}:x\left(mol\right)\end{matrix}\right.\)

=> Mỗi phần chứa \(\left\{{}\begin{matrix}KCl:a+0,5b+0,5c\left(mol\right)\\HCl_{dư}:0,5x\left(mol\right)\end{matrix}\right.\)

- Phần 1:

n_NaOH = 0,25.0,4 = 0,1 (mol)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

                0,1---->0,1

=> 0,5x = 0,1

=> x = 0,2 

- Phần 2:

\(n_{AgCl}=\dfrac{51,66}{143,5}=0,36\left(mol\right)\)

PTHH: \(KCl+AgNO_3\rightarrow AgCl\downarrow+KNO_3\)

  (a+0,5b+0,5c)-------->(a+0,5b+0,5c)

            \(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)

               0,1--------------->0,1

=> a+0,5b+0,5c + 0,1 = 0,36

=> a + 0,5b + 0,5c = 0,26 (3)

(1)(2)(3) => a = 0,2 (mol); b = 0,02 (mol); c = 0,1 (mol)

\(\left\{{}\begin{matrix}m_{K_2CO_3}=0,2.138=27,6\left(g\right)\\m_{KCl}=0,02.74,5=1,49\left(g\right)\\m_{KHCO_3}=0,1.100=10\left(mol\right)\end{matrix}\right.\)

n_HCl(bđ) = 2a + c + x = 0,7 (mol)

=> m_HCl(bd) = 0,7.36,5 = 25,55 (g)

=> \(m_{dd.HCl}=\dfrac{25,55.100}{10,52}\approx242,87\left(g\right)\)

=> \(V_{dd.HCl}=\dfrac{242,87}{1,05}\approx231,3\left(ml\right)\)

Gọi kim loại đó là R

=> CT oxit, muối sunfat là RO, RSO₄ 

Gọi \(\left\{{}\begin{matrix}n_R=a\left(mol\right)\\n_{RO}=b\left(mol\right)\\n_{RSO_4}=c\left(mol\right)\end{matrix}\right.\)

=> \(aM_R+b\left(M_R+16\right)+c\left(M_R+96\right)=14,8\) (1)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right);n_{CuSO_4}=0,2.2=0,4\left(mol\right)\)

PTHH:

\(R+H_2SO_4\rightarrow RSO_4+H_2\)

a----------------->a-------->a

\(RO+H_2SO_4\rightarrow RSO_4+H_2O\)

b-------------------->b

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ RSO_4+2NaOH\rightarrow R\left(OH\right)_2\downarrow+Na_2SO_4\)

(a + b + c)------------>(a + b + c)

\(R\left(OH\right)_2\xrightarrow[]{t^o}RO+H_2O\)

(a + b + c)->(a + b + c)

=> a = 0,2 (2)

     \(\left(a+b+c\right)\left(M_R+16\right)=14\) (3)

                          \(R+CuSO_4\rightarrow RSO_4+Cu\downarrow\)

ban đầu            0,2      0,4

p/ứ                    0,2--->0,2

sau p/ứ             0         0,2

=> \(0,2.160+\left(0,2+c\right)\left(M_R+96\right)=62\) (4)

(1),(2),(3),(4) => \(\left\{{}\begin{matrix}a=0,2\\b=0,1\\c=0,05\\M_R=24\end{matrix}\right.\)

=> R là Mg

=> Oxit kim loại là MgO

A: O₂

B: SO₂

C: SO₃

D: BaSO₃

E: Ba(HSO₃)₂

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(S+O_2\underrightarrow{t^o}SO_2\)

\(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(Cu+2H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\)

\(BaSO_3+SO_2+H_2O\rightarrow Ba\left(HSO_3\right)_2\)

\(Ba\left(HSO_3\right)_2\underrightarrow{t^o}BaSO_3+SO_2+H_2O\)

Gọi \(\left\{{}\begin{matrix}n_{O_2\left(bđ\right)}=a\left(mol\right)\\n_{SO_2\left(bđ\right)}=b\left(mol\right)\end{matrix}\right.\)

Ta có: \(\dfrac{32a+64b}{a+b}=24.2=48(g/mol)\)

=> a = b 

Giả sử a = b = 1 (mol)

=> m_{khí sau pư} = 1.32 + 1.64 = 96 (g)

Gọi số mol O₂ pư là x (mol)

PTHH:              \(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)

Bđ                         1        1

Pư:                     2x<-----x---------->2x

Sau pư:           (1-2x)    (1-x)            2x

Ta có: \(\dfrac{96}{\left(1-2x\right)+\left(1-x\right)+2x}=30.2=60\left(g/mol\right)\)

=> x = 0,4 (mol)

Sau pư thu được \(\left\{{}\begin{matrix}SO_2:0,2\left(mol\right)\\O_2:0,6\left(mol\right)\\SO_3:0,8\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,2}{0,2+0,6+0,8}.100\%=12,5\%\\\%V_{O_2}=\dfrac{0,6}{0,2+0,6+0,8}.100\%=37,5\%\\\%V_{SO_3}=\dfrac{0,8}{0,2+0,6+0,8}.100\%=50\%\end{matrix}\right.\)

\(\%O_2\left(pư\right)=\dfrac{0,4}{1}.100\%=40\%\)

\(\%SO_2\left(pư\right)=\dfrac{0,8}{1}.100\%=80\%\)