3R+8HNO₃\(\rightarrow\)3R(NO₃)₂+2NO+4H₂O

\(n_{NO}=\dfrac{6,72}{22,4}=0,3mol\)

\(n_R=\dfrac{3}{2}n_{NO}=\dfrac{3}{2}.0,3=0,45mol\)

M_R=\(\dfrac{93,15}{0,45}=207\left(Hg\right)\)

a) Al(OH)3: oxit bazơ: nhôm oxit

H2SO4: Axit: axit sunfuric

Al2(SO4)3: muối: nhôm sunfat

H2O: nước

a) Các chất trong phản ứng:

Al(OH)3 : oxit bazo: Nhôm hidroxit

H2SO4: axit : axit sunfuric

Al2(SO4)3: Muối: Nhôm sunfat

H2O: Nước

b) \(n_{Al\left(OH\right)3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

PTHH:

2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + H2O

B.đầu: 0,1________0,2_________0________0

P.ứng: 0,1_________0,15_______0,05______0,05

S.P.ứng: 0,1_________0,05________0,05_____0,05 (mol)

+) m_H2SO4(dư) = 0,05. 98 = 4,9 (gam)

+) m_Al2(SO4)3 = 0,05. 342 = 17,1 (gam)

+) m_H2O = 0,05.18 = 0,9 (gam)

a,

Ca: kim loại

CaO: oxit bazo

Ca(OH)\(_2\) : bazo

CaCl\(_2\) : Muối

HCl: axit

Cl\(_2\) : khí clo

b, \(2Ca+O_2\rightarrow2CaO\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(Ca\left(OH\right)_2+CuCl_2\rightarrow Cu\left(OH\right)_2+CaCl_2\)

\(CaCl_2+H_2\rightarrow Ca+2HCl\)

\(4HCl+MnO_2\rightarrow MnCl_2+Cl_2+2H_2O\)

a, Ca: kim loại

CaO: oxit bazơ

Ca(OH)₂ : dd Bazơ

CaCl₂ :Muối

HCl: Axit

Cl₂ : Phi kim

b, 2Ca +O₂ --^to--> 2CaO

CaO + H₂O ----> Ca(OH)₂

Ca(OH)₂ + MgCl₂ ----> CaCl₂ +Mg(OH)₂

CaCl₂ +H₂SO₄ -----> CaSO₄ + 2HCl

6HCl + 2Al -------> 2AlCl₃ +3H₂

xin lỗi, mình đính chính phương trình cuối :

2KnO4 +16 HCl ---^_to-> 2KCl + 5Cl₂ + 8H₂O

Bài 1:

\(CH4+2O2-->CO2+2H2O\)

\(n_{CH4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{O2}=2n_{CH4}=0,5\left(mol\right)\)

\(V_{O2}=0,.22,4=11,2\left(l\right)\)

\(V_{kk}=5V_{O2}=11,2.5=56\left(l\right)\)

Bài 2:

\(m_{CaCO3}=280.89,29\%\approx250\left(tấn\right)\)

\(CaCO3-->CaO+CO2\)

\(m_{CO2}=m_{CaCO3}-m_{CaO}=250-140=110\left(tấn\right)\)

Bài 3:

Có 4% tạp chất k cháy =>96% C

\(m_C=1.96\%=0,96\left(kg\right)=960\left(g\right)\)

\(n_C=\frac{960}{12}=80\left(mol\right)\)

\(C+O2-->CO2\)

\(n_{O2}=n_C=80\left(mol\right)\)

\(V_{O2}=80.22,4=1792\left(l\right)\)

Bài 4:

2M + 2H2O -----> 2MOH + H2

\(n_{H2}=\frac{1,68}{22,4}=0,075\left(mol\right)\)

\(n_M=2n_{H2}=0,15\left(mol\right)\)

\(M_M=\frac{5,85}{0,15}=39\left(K\right)\)

Vậy M có NTK là 39

Bài 5:

M + 2HCl -----> MCl2 + H2

\(n_{H2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)

\(n_M=n_{H2}=0,025\left(mol\right)\)

\(M_M=\frac{1,4}{0,025}=56\left(Fe\right)\)

vậy M là Fe

Bài 11:

Đề là 15,6 đúng hơn nha bạn

2R + 2nHCl ----> 2RCln + nH2

\(n_{H2}=\frac{5,376}{22,4}=0,24\left(mol\right)\)

\(n_R=\frac{2}{n}n_{H2}=\frac{0,48}{n}\left(mol\right)\)

\(M_R=15,6:\frac{0,48}{n}=32,5n\)

\(n=2\Rightarrow M_M=65\left(Zn\right)\)

Vậy M là Zn

a) Zn+2HCl---->ZnCl2+H2

b) n H2=4,48/22,4=0,2(mol)

n Zn=n ZnCl2=n H2=0,2(mol)

m Zn=0,2.65=13(g)

m Zncl2=0,2.136=27,2(g)

c) CuO+H2---->Cu+H2O

n CuO=24/80=0,3(mol)

--->CuO dư

n CuO=n H2=0,2(mol)

n CuO dư=0,3-0,2=0,1(g)

m CuO dư=0,1.80=8(g)

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

0,2____________0,2____0,2

\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

\(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{ZnCl2}=0,2.\left(65+71\right)=27,2\left(g\right)\)

\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)

Ban đầu :0,3____0,2____________

Phứng: 0,2______0,2__________

Sau phứng :0,1___0___________

\(n_{CuO}=\frac{24}{80}=0,3\left(mol\right)\)

Vậy CuO dư

\(m_{CuO_{du}}=0,1.80=8\left(g\right)\)

ta có n_Fe3O4= 185,6/ 232= 0,8( mol)

PTPU

Fe₃O₄+ 4CO---to---> 3Fe+ 4CO₂ (1)

_{..0,8mol-> 3,2mol}

theo gt n_CO= 44,8/ 22,4= 2( mol)< 3,2mol---> CO hết, Fe3O4 dư

theo(1) ta có n_Fe= 3/4 n_CO= 3/4x 2= 1,5( mol)

=> m_Fe= 1,5x 56= 84( g)

theo(1) ta có n_CO2= n_CO= 2( mol)

=> V_CO2= 2x 22,4= 44,8( l)

b) PTPU

CO₂+ Ca(OH)₂---> CaCO₃+ H₂O (2)

theo(2) ta có n_CaCO3= n_CO2= 2(mol)

=> m_CaCO3= 2x 100= 200( g)