$a) MnO_2 + 4HCl \xrightarrow{t^o} MnCl_2 + Cl_2 + 2H_2O$
$Cl_2 + Cu \xrightarrow{t^o} CuCl_2$

$CuCl_2 + 2NaOH \to Cu(OH)_2 + 2NaCl$
$2NaCl + 2H_2O \xrightarrow{đpdd, cmn} 2NaOH + H_2 + Cl_2$
$2NaOH + Cl_2 \to NaCl + NaClO + H_2O$
$NaClO + HCl \to NaCl + HClO$

b)

$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$
$Cl_2 + H_2 \xrightarrow{ánh\ sáng} 2HCl$
$HCl + KOH \to KCl + H_2O$

$2KCl + 2H_2O \xrightarrow{đpdd, cmn} 2KOH + H_2 + Cl_2$
$3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3$

$FeCl_3 + 3AgNO_3 \to Fe(NO_3)_3 + 3AgCl$

(1) \(2NaCl+2H_2O\underrightarrow{đpcmn}2NaOH+H_2+Cl_2\)

(2) \(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)

(3) \(H_2+Br_2\underrightarrow{t^o}2HBr\)

(4) \(HBr+AgNO_3\rightarrow AgBr\downarrow+HNO_3\)

\(CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ CO_2 + 2NaOH \to Na_2CO_3 + H_2O\\ Na_2CO_3 + Ba(OH)_2 \to BaCO_3 + 2NaOH\)

\(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)

\(CO_2+Na_2O\rightarrow Na_2CO_3\)

\(Na_2CO_3+Ba\left(OH\right)_2\rightarrow2NaOH+BaCO_3\downarrow\)

CaCO3 -to-> CaO + CO2

CO2 + 2 NaOH -> Na2CO3 + H2O

Na2CO3 + Ca(OH)2 -> CaCO3 + 2 NaOH

Hoàn thành các phương trình phản ứng xảy ra (nếu có, ghi rõ điều kiện)

1) 3Cl₂ + 2Fe →to->...2FeCl3........

2) 2Cl₂ + 2H₂O → .....4HCl + O2.......

3) Cl₂ + 2NaOH → ....NaClO + NaCl + H20.....

4) Fe + .....2HCl..... → FeCl₂ + ......H2.......

5) 3HCl + Fe(OH)₃→......FeCl3 + 3H20.............

6) KCl + AgNO₃→........AgCl + KNO3................

7) 2HCl + CaCO₃ →.........CaCl2 + H2CO3..............

ai hộ e với em cần gấp ạ

Câu 1 : 

\(a.Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{t^0}}FeCl_3\)

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

\(c.2H_2S+3O_{2\left(dư\right)}\underrightarrow{^{t^0}}2H_2O+2SO_2\)

\(d.2FeS+10H_2SO_{4\left(đ\right)}\underrightarrow{^{t^0}}Fe_2\left(SO_4\right)_3+9SO_2+10H_2O\)

Câu 2 : 

Trích mẫu thử

Cho quỳ tím vào các mẫu thử

- mẫu thử nào làm quỳ tím hóa đỏ là H₂SO₄

- mẫu thử nào làm quỳ tím hóa xanh là Na₂S

Cho dung dịch Bari clorua vào mẫu thử còn :

- mẫu thử nào tạo kết tủa trắng là MgSO₄

\(BaCl_2 +MgSO_4 \to BaSO_4 + MgCl_2\)

- mẫu thử nào không hiện tượng gì là KCl

Câu 3 : 

\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)

\(m_A=56a+24b=13.2\left(g\right)\left(1\right)\)

\(n_{SO_2}=\dfrac{9.52}{22.4}=0.425\left(mol\right)\)

\(n_{H_2SO_4\left(bđ\right)}=\dfrac{100\cdot98}{100\cdot98}=1\left(mol\right)\)

\(a.\)

\(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{^{t^0}}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

\(\)\(Mg+2H_2SO_{4\left(đ\right)}\underrightarrow{^{t^0}}MgSO_4+SO_2+2H_2O\)

\(n_{SO_2}=1.5a+b=0.425\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.2\)

\(\%m_{Fe}=\dfrac{0.15\cdot56}{13.2}\cdot100\%=63.64\%\)

\(\%m_{Mg}=36.36\%\)

\(b.\)

\(n_{H_2SO_4\left(dư\right)}=1-0.425\cdot2=0.15\left(mol\right)\)

\(m_{\text{B}}=m_A+m_{dd_{H_2SO_4}}-m_{SO_2}=13.2+100-0.425\cdot64=86\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0.075\cdot400}{86}\cdot100\%=34.88\%\)

\(C\%_{MgSO_4}=\dfrac{0.2\cdot120}{86}\cdot100\%=27.91\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0.15\cdot98}{86}\cdot100\%=17.09\%\)

\(c.\)

\(n_{NaOH}=\dfrac{119\cdot20}{40\cdot100}=0.595\left(mol\right)\)

\(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0.595}{0.425}=1.4\) \(\Rightarrow\text{Tạo ra 2 muối}\)

\(n_{Na_2SO_3}=x\left(mol\right),n_{NaHSO_3}=y\left(mol\right)\)

\(\left\{{}\begin{matrix}2x+y=0.595\\x+y=0.425\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0.17\\y=0.255\end{matrix}\right.\)

\(m_{Na_2SO_3}=0.17\cdot126=21.42\left(g\right)\)

\(m_{NaHSO_3}=0.255\cdot104=26.52\left(g\right)\)

\(d.\)

\(n_{Ba\left(OH\right)_2}=0.2\cdot0.35=0.07\left(mol\right)\)

\(n_{BaCl_2}=0.2\left(mol\right)\)

\(OH^-+H^+\rightarrow H_2O\)

\(0.07......0.07\)

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}+n_{BaCl_2}=0.07+0.2=0.27\left(mol\right)\)

\(n_{SO_4^{2-}}=3n_{Fe_2\left(SO_4\right)_3}+n_{MgSO_4}+n_{H_2SO_4\left(\text{còn lại}\right)}=3\cdot0.075+0.2+0.15-0.14=0.435\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

\(0.27.........0.27.........0.27\)

\(\Rightarrow SO_4^{2-}\text{dư}\)

\(m_{BaSO_4}=0.27\cdot233=62.91\left(g\right)\)

a)

2H₂O-đp>2H₂+O₂

b) SO4 ??

c)

2KClO3-to>2KCl+3O₂

S+O₂-to>SO₂

SO₂+O₂-to, V2O5>SO₃

SO₃+H₂O->H₂SO₄

2Cu+O₂-to>2CuO

CuO+H₂SO₄->CuSO₄+H₂O

d) KClO₃ + 6HCl → 3Cl₂ + KCl + 3H₂O

Cl₂+Cu-to>CuCl₂

2H2O -> (điện phân) 2H2 + O2

2KClO3 -> (t°, MnO2) 2KCl + 3O2

S + O2 -> (t°) SO2

2SO2 + O2 -> (t°, V2O5) 2SO3

SO3 + H2O -> H2SO4

Cu + H2SO4 (đặc nóng) -> CuSO4 + H2

2Cu + O2 -> (t°) 2CuO 

CuO + 2HCl -> CuCl2 + H2O

a)

2H₂O-đp>2H₂+O₂

b) SO4 ??

c) ko có gốc S thì sao SO4 đc

d) KClO₃ + 6HCl → 3Cl₂ + KCl + 3H₂O

Cl₂+Cu-to>CuCl₂