3Fe+2O₂-to>Fe₃O₄

0,225---0,15---0,075

n _O2= \(\dfrac{3,7185}{24,79}\)=0,15 mol

=>m _Fe=0,225.56=12,6g

=>m _Fe3O4=0,075.232=17,4g

Gọi oxit kim loại cần tìm là RO

\(n_{RO}=\dfrac{5,6}{M_R+16}\left(mol\right)\)

PTHH: RO + H₂O --> R(OH)₂ 

         \(\dfrac{5,6}{M_R+16}\)---------->\(\dfrac{5,6}{M_R+16}\)

=> \(M_{R\left(OH\right)_2}=M_R+34=\dfrac{7,4}{\dfrac{5,6}{M_R+16}}=\dfrac{37}{28}\left(M_R+16\right)\left(g/mol\right)\)

=> M_R = 40 (g/mol)

=> Kim loại là Ca

\(CO_2\) -> oxit axit -> cacbonic

\(Fe_2O_3\) -> oxit bazo -> sắt (III) oxit

\(N_2O_5\) -> oxit axit -> đinito pentaoxit

\(MgO\) -> oxit bazo -> magie oxit

CO₂ : oxit axit : Cacbondioxit

Fe₂O₃ : oxit bazo : Sắt 3 oxit

N₂O₅ : oxit axit : Đinitopentaoxit

MgO :oxit bazo :Magie oxit

50 CTHH là sao em? Hay PTHH? Mà 50 nhiều thế

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_{P_2O_5}=\dfrac{2}{4}.n_P=\dfrac{2.0,2}{4}=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\)

\(n_{CH_4}=\dfrac{5,6}{22,4}=0,25mol\)

\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

0,25     0,5       0,25      0,5

\(V_{O_2}=0,5\cdot22,4=11,2l\)

\(V_{CO_2}=0,25\cdot22,4=5,6l\)

\(m_{H_2O}=0,5\cdot18=9g\)

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

        0,2-->0,4--------------->0,2

=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

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

0,2      0,4                        0,2

\(m_{HCl}=0,4\cdot36,5=14,6g\)

\(V_{H_2}=0,2\cdot22,4=4,48l\)

\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)

\(n_{H_2SO_4}=\dfrac{7,35}{98}=0,075mol\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{8,55}{342}=0,025mol\)

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

0,05     0,075          0,025           0,075

\(a=m_{H_2}=0,075\cdot2=0,15g\)

\(V_{H_2}=0,075\cdot22,4=1,68l\)

\(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)