a) H₂

PTK = 2.1 = 2 (đvC)

b) O₂

PTK = 16.2 = 32 (đvC)

c) Na₂SO₄

PTK = 23.2 + 32.1 + 16.4 = 142 (đvC)

d) AgCl

PTK = 108.1 + 35,5.1 = 143,5 (đvC)

e) Ca(NO₃)₂

PTK = 40.1 + (14.1 + 16,3).2 = 164 (đvC)

g) Mg₃(PO₄)₂

PTK = 24.3 + (31.1 + 16.4).2 = 262 (đvC)

a) 

TH1: Gọi \(\left(n_{Al};n_{Cu};n_{Mg}\right)=\left(a;b;c\right)\)

=> 27a + 64b + 24c = 13,98 (1)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo PTHH: \(n_{H_2}=1,5a+c=0,3\) (2)

TH2: Gọi \(\left(n_{Al};n_{Cu};n_{Mg}\right)=\left(ak;bk;ck\right)\)

=> ak + bk + ck = 0,666 (*)

PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

            \(2Cu+O_2\underrightarrow{t^o}2CuO\)

            \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,5ak\left(mol\right)\\n_{CuO}=bk\left(mol\right)\\n_{MgO}=ck\left(mol\right)\end{matrix}\right.\)

=> 51ak + 80bk + 40ck = 37,26 (**)

(*)(**) => \(\dfrac{a+b+c}{51a+80b+40c}=\dfrac{37}{2070}\)

=> 183a - 890b + 590c = 0 (3)

(1)(2)(3) => a = 0,1 (mol); b = 0,12 (mol); c = 0,15 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{13,98}.100\%=19,31\%\\\%m_{Cu}=\dfrac{0,12.64}{13,98}.100\%=54,94\%\\\%m_{Mg}=\dfrac{0,15.24}{13,98}.100\%=25,75\%\end{matrix}\right.\)

b) 0,814 mol X chứa \(\left\{{}\begin{matrix}n_{Al}=0,22\left(mol\right)\\n_{Cu}=0,264\left(mol\right)\\n_{Mg}=0,33\left(mol\right)\end{matrix}\right.\)

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

           0,22-------------------->0,11---->0,33

            \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

           0,33----------------->0,33---->0,33

=> \(V_{H_2}=\left(0,33+0,33\right).22,4=14,784\left(l\right)\)

m_muối = 0,11.342 + 0,33.120 = 77,22 (g)

Câu a) bn hỏi rồi nhé

b)

\(n_{O_2}=\dfrac{4,8}{22,4}=\dfrac{3}{14}\left(mol\right)\)

Gọi số mol Fe, Cu là a, b (mol)

PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

              a--->\(\dfrac{2}{3}a\)

            \(2Cu+O_2\underrightarrow{t^o}2CuO\)

              b---->0,5b

=> \(\left\{{}\begin{matrix}\dfrac{2}{3}a+0,5b=\dfrac{3}{14}\\\dfrac{2}{3}a=2.0,5b=b\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=\dfrac{3}{14}\left(mol\right)\\b=\dfrac{1}{7}\left(mol\right)\end{matrix}\right.\) => m_hh = \(\dfrac{3}{14}.56+\dfrac{1}{7}.64=\dfrac{148}{7}\left(g\right)\)

a)

PTHH: \(Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\)

\(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\) => n_{O(mất đi)} = 0,4 (mol)

\(n_{Fe\left(A\right)}=\dfrac{28,4.59,155\%}{56}=0,3\left(mol\right)\)

Xét n_Fe(A) : n_{O(mất đi)} = 0,3 : 0,4 = 3 : 4

=> CTHH: Fe₃O₄

b)

Do sau pư thu được muối sunfat => A là muối sunfat

B là M(NO₃)_m

\(n_{M\left(NO_3\right)_m}=\dfrac{50.10,44\%}{M_M+62m}\left(mol\right)\)

Bảo toàn M: 

 \(n_{M\left(NO_3\right)_m}=2.n_{M_2\left(SO_4\right)_m}\)

=> \(\dfrac{5,22}{M_M+62m}\) = \(2.\dfrac{4,66}{2.M_M+96m}\)

=> 10,44.M_M + 501,12.m = 9,32.M_M + 577,84m

=> M_M = 68,5m (g/mol)

Xét m = 2 thỏa mãn => M_M = 137 (g/mol)

=> M là Ba

B là Ba(NO₃)₂

\(n_{Ba\left(NO_3\right)_2}=\) 0,02 (mol)

m_A = 200.1,36% = 2,72 (g)

TH1: A là X(HSO₄)_n

PTHH: \(X\left(HSO_4\right)_n+nBa\left(NO_3\right)_2\rightarrow nBaSO_4+X\left(NO_3\right)_n+nHNO_3\)

                \(\dfrac{0,02}{n}\)<--------0,02

=> \(M_{X\left(HSO_4\right)_n}=\dfrac{2,72}{\dfrac{0,02}{n}}=136n\left(g/mol\right)\Rightarrow M_X=39n\left(g/mol\right)\)

Xét n = 1 thỏa mãn => M_X = 39 (g/mol) => X là K

A là KHSO₄

TH2: B là X₂(SO₄)_n

PTHH: \(X_2\left(SO_4\right)_n+nBa\left(NO_3\right)_2\rightarrow nBaSO_4+2X\left(NO_3\right)_n\)

               \(\dfrac{0,02}{n}\)<----------0,02

=> \(M_{X_2\left(SO_4\right)_n}=\dfrac{2,72}{\dfrac{0,02}{n}}=136n\left(g/mol\right)\Rightarrow M_X=20n\left(g/mol\right)\)

Xét n = 2 thỏa mãn => M_X = 40 (g/mol)

=> X là Ca (Loại do CaSO₄ kết tủa)

a)

Gọi số mol Fe, Zn là a, b (mol)

=> 56a + 65b = 8,2 (1)

PTHH: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

              a--------------->a

           \(Zn+Cl_2\underrightarrow{t^o}ZnCl_2\)

             b------------->b

=> 162,5a + 136b = 21,69 (2)

(1)(2) => a = 0,1 (mol); b = 0,04 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{8,2}.100\%=68,3\%\\\%m_{Zn}=\dfrac{0,04.65}{8,2}.100\%=31,7\%\end{matrix}\right.\)

b)

Ta có: \(\dfrac{n_{Fe}}{n_{Zn}}=\dfrac{0,1}{0,04}=2,5\)

Mà n_Fe + n_Zn = 0,07 

=> n_Fe = 0,05 (mol); n_Zn = 0,02 (mol)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

          0,05--------------------->0,05

            \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

           0,02----------------------->0,02

=> V = (0,05 + 0,02).22,4 = 1,568 (l)

a) \(n_{O_2}=\dfrac{4,8}{22,4}=\dfrac{3}{14}\left(mol\right)\)

PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

                  \(\dfrac{3}{7}\)<------------------------------\(\dfrac{3}{14}\)

=> \(m_{KMnO_4\left(pư\right)}=\dfrac{3}{7}.158=\dfrac{474}{7}\left(g\right)\)

=> \(m_{KMnO_4\left(TT\right)}=\dfrac{474}{7}:80\%=\dfrac{1185}{14}\left(g\right)\)

a)

Gọi số mol Al, Mg trong mỗi phần là a, b (mol)

Phần 1:

\(n_{H_2}=\dfrac{5,152}{22,4}=0,23\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo PTHH: \(n_{H_2}=1,5a+b=\) 0,23 (1)

Phần 2:

PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

            \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,5a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\)

=> 102.0,5a + 40b = 8,3

=> 51a + 40b = 8,3 (2)

(1)(2) => a = 0,1 (mol); b = 0,08 (mol)

=> m = 2.(0,1.27 + 0,08.24) = 9,24 (g)

b) Theo PTHH: \(n_{O_2}=0,75a+0,5b\) = 0,115 (mol)

=> \(V_{O_2}=0,115.22,4=2,576\left(l\right)\)

CTHH: CaAO₄

Ta có: PTK = 1,9.71 \(\approx135\left(đvC\right)\)

=> 40 + NTK_A  + 16.4 = 135

=> NTK_A = 31 (đvC)

=> A là P(Photpho)

Ta có: \(\%Fe=\dfrac{56}{56+35,5x}.100\%=34,46\%\) => x = 3

CTHH: Fe^xCl₃^I

Theo quy tắc hóa trị, ta có: x.1 = 3.I

=> x = III

Vậy Fe có hóa trị III

CTHH: A(NO₃)₂

\(M_{A\left(NO_3\right)_2}=\dfrac{51,8}{0,35}=148\left(g/mol\right)\)

=> M_A = 24 (g/mol)

=> A là Mg (Magie)