`4P + 5O_2` $\xrightarrow{t^o}$ `2P_2 O_5`

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

`n_P=[3,1]/31=0,1(mol)`

`=>m_[P_2 O_5]=0,05.142=7,1(g)`

- Phương pháp điều chế: cho các kim loại (Al, Mg, Fe, Zn,...) tác dụng với dd axit (HCl, H₂SO₄ loãng,...)

- Cách thu: ngửa bình

- Khác với cách thu oxi: Thu oxi thì phải úp bình

\(m_{dd}=\dfrac{50.100}{20}=250\left(g\right)\\ \rightarrow m_{H_2O}=250-50=200\left(g\right)\)

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,1`  `0,2`               `0,1`     `0,1`          `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`b)m_[HCl]=0,2.36,5=7,3(g)`

`c)`

`@m_[ZnCl_2]=0,1.136=13,6(g)`

`@V_[H_2]=0,1.22,4=2,24(l)`

\(n_{CuSO_4}=\dfrac{400}{160}=2,5\left(mol\right)\\ C_M=\dfrac{2,5}{4}=0,625M\)

\(n_{CuSO_4}=\dfrac{400}{160}=2,5\left(mol\right)\\ C_{M\left(CuSO_4\right)}=\dfrac{2,5}{4}=0,625M\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1     0,2           0,1          0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ m_{ZnCl_2}=1360,1=13,6\left(g\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Zn+2HCl->ZnCl2+H2

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

n Zn=0,1 mol

=>m HCl=0,2.36,5=7,3g

=>m ZnCl2=0,1.136=13,6g

=>VH2=0,1.22,4=2,24l

\(n_{H_2SO_4}=0,5.0,15=0,075\left(mol\right)\)

0,15 la j vay

Zn+2HCl->ZnCl2+H2

Zn+H2SO4->ZnSO4+H2

Fe+2HCl->FeCl2+H2

Fe+H2SO4->FeSO4+H2

\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

nHCl=0,5.5=2,5(mol)

mdd HCl=\(\dfrac{2,5.36,5}{36,5\%}\)%=250(gam)
Vdd HCl=\(\dfrac{250}{1,19}\)=210(ml)

\(\%m_H=\dfrac{2}{18}.100\%=11,1\%\\ m_H=3,6.11,1\%=0,3996\left(g\right)\)