2KClO₃ --t^o-->2KCl + 3O₂

4Na + O₂ --t^o--> 2Na₂O

Na₂O + H₂O --> 2NaOH

\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)

Áp dụng ĐLBTNT: \(m_O=12-9,6=2,4\left(g\right)\)

\(\rightarrow n_O=\dfrac{2,4}{16}=0,15\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Theo pthh: \(\left\{{}\begin{matrix}n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_O=\dfrac{1}{3}.0,15=0,05\left(mol\right)\\n_{H_2\left(pư\right)}=n_O=0,15\left(mol\right)\end{matrix}\right.\)

LTL: \(\dfrac{0,3}{3}>0,075\) => hiệu suất pư tính theo Fe₂O₃

=> \(H=\dfrac{0,05}{0,075}.100\%=66,67\%\)

đề cóa saii hongg bạn?:)

CaO - oxit - canxi oxit 
Mg(OH)2 - bazo - magie hidroxit 
SO3 - oxit - lưu huỳnh trioxit 
HCl - axit - axit clohidric 
Fe2O3 - oxit - sắt (III) oxit 
Al(OH)3 - bazo -nhôm hidroxit 
P2O3 - oxit - đi phot pho tri oxit 
H2SO4 -axit - axit sunfuric 
Ca(HCO3)2 - muối - canxi hidro cacbonat 
HNO3 - axit - axit nitric 
CO2 - oxit -cacbon dioxit 
CuO - oxit - đồng(II) oxit  
2 
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ K_2O+H_2O\rightarrow2KOH\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ SO_3+H_2O\rightarrow H_2SO_4\\ CO_2+H_2O\rightarrow H_2CO_3\\ Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ N_2O_5+H_2O\rightarrow2HNO_3\)

CaO - oxit - canxi oxit 
Mg(OH)2 - bazo - magie hidroxit 
SO3 - oxit - lưu huỳnh trioxit 
HCl - axit - axit clohidric 
Fe2O3 - oxit - sắt (III) oxit 
Al(OH)3 - bazo -nhôm hidroxit 
P2O3 - oxit - đi phot pho tri oxit 
H2SO4 -axit - axit sunfuric 
Ca(HCO3)2 - muối - canxi hidro cacbonat 
HNO3 - axit - axit nitric 
CO2 - oxit -cacbon dioxit 
CuO - oxit - đồng(II) oxit  
2 
\(Ca+H_2O\rightarrow Ca\left(OH\right)_2\\ K_2O+H_2O\rightarrow2KOH\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ SO_3+H_2O\rightarrow H_2SO_4\\ CO_2+H_2O\rightarrow H_2CO_3\\ Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ N_2O_5+H_2O\rightarrow2HNO_3\)
 

\(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PTHH:

2K + 2H₂O ---> 2KOH + H₂

0,2<---------------0,2<----0,1

=> \(\left\{{}\begin{matrix}m_K=0,2.39=7,8\left(g\right)\\m_{K_2O}=12,5-7,8=4,7\left(g\right)\end{matrix}\right.\)

\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)

PTHH: K₂O + H₂O ---> 2KOH

            0,05-------------->0,1

=> m_KOH = (0,2 + 0,1).56 = 16,8 (g)

2K+2H2O->2KOH+H2

0,2--------------0,2-------0,1mol

K2O+H2O->2KOH

0,05--------------0,1

n H2=0,1 mol

=>m K=0,2.39=7,8g

=>m K2O=4,7g=>n K2O=0,05 mol

=>m bazo=0,3.56=16,8g

bạn ý bảo đổi 12,5 = 7,8 
\(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ pthh:X+H_2O\rightarrow XOH+\dfrac{1}{2}H_2O\)
         0,2                                0,1 
\(M_X=\dfrac{7,8}{0,2}=39\left(\dfrac{g}{mol}\right)\) 
có X hóa trị I  
=> X là Kali (K)

2X+2H2O->2XOH+H2

ta có :\(\dfrac{12,5}{X}=\dfrac{2,479}{24,79}.2\)

=>X=62,5??

2Na+2H2O->2NaOH+H2

0,3------------------------------0,15

n H2=0,15 mol

=>m Na=0,3.23=6,9g

=>m Cu=10-6,9=3,1g

->quỳ chuyển xanh , vì có dd kiềm NaOH

\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\) 
           0,3                                 0,15 
          \(Cu+H_2O\) -/-> 
\(m_{Na}=0,3.23=6,9g\\ \Rightarrow m_{Cu}=10-6,9=3,1g\) 
dd sau pư làm QT chuyển xanh vì NaOH là bazo 

\(n_K=\dfrac{3,9}{39}=0,1mol\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

0,1                        0,1      0,05 ( mol )

\(m_{KOH}=0,1.56=5,6g\)

\(V_{H_2}=0,05.22,4=1,12l\)

mik cảm ơn

1 tấn =1000kg 

=>có 95%C=>950kg

C+O2-to>CO2

79,167--79,167

n C=\(\dfrac{950}{12}\)=79,167 mol

=>mO2=79,167.32=2533,33kg

2Al+3H2SO4+Al2(SO4)3+3H2

0,2-----0,3----------------------0,3

H2+ZnO-to>Zn+H2O

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

=>m H2SO4=0,3.98=29,4g

=>C% =\(\dfrac{29,4}{150}100\)=19,6%

=>m Zn=0,3.65=19,5g

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

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

0,15    0,3                         0,15  ( mol )

\(m_{Fe}=0,15.56=8,4g\)

\(m_{ddHCl}=\dfrac{0,3.36,5}{20\%}=54,75g\)

chị uii sửa thêm chỗ m HCl nữa chị:<