a)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)

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

b) Giả sử số mol FeO, Fe₃O₄, Fe₂O₃ trong mỗi phần là a, b, c (mol)

=> 72a + 232b + 160c = 78,4 (g) (*)

- Phần 1:

Theo PTHH: \(\left\{{}\begin{matrix}n_{FeCl_2}=a+b\left(mol\right)\\n_{FeCl_3}=2b+2c\left(mol\right)\end{matrix}\right.\)

=> 127(a + b) + 162,5(2b + 2c) = 155,4

=> 127a + 452b + 325c = 155,4 (**)

Và \(n_{H_2O}\) = 0,5.n_HCl(pư) = 0,5(2a + 8b + 6c)

= a + 4b + 3c (mol) (1)

- Phần 2:

Gọi \(\left\{{}\begin{matrix}n_{HCl}=x\left(mol\right)\\n_{H_2SO_4}=y\left(mol\right)\end{matrix}\right.\)

Theo PTHH: \(n_{H_2O}=0,5.n_{HCl}+n_{H_2SO_4}=0,5x+y\left(mol\right)\) (2)

(1)(2) => a + 4b + 3c = 0,5x + y 

Theo ĐLBTKL: 

\(m_{oxit}+m_{HCl}+m_{H_2SO_4}=m_{muối}+m_{H_2O}\)

=> 78,4 + 36,5x + 98y = 167,9 + 18(0,5x + y)

=> 27,5x + 80y = 89,5 (3)

Trừ theo vế (**),(*) => 55a + 220b + 165c = 77

=> a + 4b + 3c = 1,4 = 0,5x + y (4)

(3)(4) => x = 1,8 (mol)

\(\)

a) \(n_{HCl}=0,3.2=0,6\left(mol\right)\)

PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

               0,1<-----0,6------->0,2

=> \(m_{Fe_2O_3}=0,1.160=16\left(g\right)\)

b) \(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

c) m_{dd HCl} = 300.1,05 = 315 (g)

=> m_{dd sau pư} = 315 + 16 = 331 (g)

\(C\%_{FeCl_3}=\dfrac{32,5}{331}.100\%=9,82\%\)

a) \(n_{H_2SO_4\left(bđ\right)}=0,3.1=0,3\left(mol\right)\Rightarrow n_{H_2SO_4\left(pư\right)}=\dfrac{0,3.100}{120}=0,25\left(mol\right)\)

Oxit có CTHH là RO

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

            0,25<--0,25---->0,25

=> \(M_{RO}=\dfrac{14}{0,25}=56\left(g/mol\right)\Rightarrow M_R=40\left(g/mol\right)\)

=> R là Ca

CTHH: CaO

b) Tinh thể có CTHH là CaSO₄.nH₂O

\(n_{CaSO_4.nH_2O}=n_{CaSO_4}=0,25\left(mol\right)\)

=> \(M_{CaSO_4.nH_2O}=\dfrac{43}{0,25}=172\left(g/mol\right)\)

=> n = 2

CTHH: CaSO₄.2H₂O

Gọi số mol Al, Mg là a, b (mol)

=> 27a + 24b = 15 (1)

\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

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

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

Theo PTHH: \(n_{H_2}=1,5a+b=0,7\) (2)

(1)(2) => a = 0,2 (mol); b = 0,4 (mol)

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{15}.100\%=36\%\\\%Mg=\dfrac{0,4.24}{15}.100\%=64\%\end{matrix}\right.\)

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

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

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

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

\(n_{H_2SO_4}=1.0,8=0,8\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=n_{H_2SO_4}=0,8\left(mol\right)\)

Theo ĐLBTKL: \(m_{oxit}+m_{H_2SO_4}=m_{Muối}+m_{H_2O}\)

=> a = 38,3 + 0,8.98 - 0,8.18 = 102,3 (g)

1)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

n_HCl = 0,5.1,5 = 0,75 (mol)

PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,75}{6}\) => Fe₂O₃ hết, HCl dư

PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

                0,1----->0,6------>0,2

=> \(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,75-0,6}{0,5}=0,3M\\C_{M\left(FeCl_3\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)

2)

m_{dd HCl} = 500.1,19 = 595 (g)

m_{dd sau pư} = 16 + 595 = 611 (g)

\(C\%_{FeCl_3}=\dfrac{0,2.162,5}{611}.100\%=5,32\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{0,15.36,5}{611}.100\%=0,896\%\)

(A): Cu, CuO

(B) H₂SO₄, CuSO₄

(C): SO₂

(D): K₂SO₃, KHSO₃

(E) Cu(OH)₂

(F): Cu

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

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

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

\(SO_2+KOH\rightarrow KHSO_3\)

\(2KOH+SO_2\rightarrow K_2SO_3+H_2O\)

\(K_2SO_3+BaCl_2\rightarrow BaSO_3+2KCl\)

\(2KHSO_3+2NaOH\rightarrow K_2SO_3+Na_2SO_3+2H_2O\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

a) M_A = 21.2 = 42 (g/mol)

=> 14n = 42 (g/mol)

=> n = 3

CTPT: C₃H₆

b) \(n_{C_3H_6}=n_1=\dfrac{5,6.1,6}{0,082\left(0+273\right)}\approx0,4\left(mol\right)\)

\(n_2=\dfrac{5,6.5,2}{0,082\left(0+273\right)}\approx1,3\left(mol\right)\)

=> \(n_{O_2}=1,3-0,4=0,9\left(mol\right)\)

PTHH: \(2C_3H_6+9O_2\underrightarrow{t^o}6CO_2+6H_2O\)

Xét tỉ lệ: \(\dfrac{0,4}{2}>\dfrac{0,9}{9}\) => C₃H₆ dư, O₂ dư

PTHH: \(2C_3H_6+9O_2\underrightarrow{t^o}6CO_2+6H_2O\)

               0,2<----0,9---->0,6--->0,6

B gồm \(\left\{{}\begin{matrix}CO_2:0,6\left(mol\right)\\C_3H_6=0,4-0,2=0,2\left(mol\right)\end{matrix}\right.\)

=> n₃ = 0,6 + 0,2 = 0,8 (mol)

=> \(p_3=\dfrac{0,8.0,082.\left(0+273\right)}{5,6}=3,198\left(atm\right)\)

thì bn có \(n_{N_2\left(pư\right)}=0,5\left(mol\right)\) ấy

theo PTHH \(\left\{{}\begin{matrix}n_{H_2\left(pư\right)}=1,5\left(mol\right)\\n_{NH_3\left(tạo.ra\right)}=1\left(mol\right)\end{matrix}\right.\)

=> Khi đạt cân bằng \(\left\{{}\begin{matrix}n_{N_2}=2-0,5=1,5\left(mol\right)\\n_{H_2}=8-1.5=6,5\left(mol\right)\\n_{NH_3}=1\left(mol\right)\end{matrix}\right.\)