Hóa học

Có: \(\left\{{}\begin{matrix}m_{Mg}=95,25,26\%=24\left(g\right)\\m_{Cl}=95-24=71\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\\n_{Cl}=\dfrac{71}{35,5}=2\left(mol\right)\end{matrix}\right.\)

`=>A: MgCl_2`

1) \(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{31,25}{250}=0,125\left(mol\right)\)

`=>` \(C_{M\left(CuSO_4\right)}=\dfrac{0,125}{0,1}=1,25M\Rightarrow\left\{{}\begin{matrix}\left[Cu^{2+}\right]=1,25M\\\left[SO_4^{2-}\right]=1,25M\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}n_{Na_2CO_3}=\dfrac{57,2}{286}+0,4.0,1=0,24\left(mol\right)\\V_{dd}=400\left(ml\right)=0,4\left(l\right)\end{matrix}\right.\)

`=>` \(C_{M\left(Na_2CO_3\right)}=\dfrac{0,24}{0,4}=0,6M\Rightarrow\left\{{}\begin{matrix}\left[Na^+\right]=0,6.2=1,2M\\\left[CO_3^{2-}\right]=0,6M\end{matrix}\right.\)

Áp dụng đ/l bảo toàn nguyên tố có: `n_O=y.n_[Fe_x O_y]`

    Hay `n_O=y.n_X`

     `=>n_X=[n_O]/y=[0,1]/y(mol)`

`Fe + H_2 SO_4 -> FeSO_4 + H_2 \uparrow`

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

`n_[Fe]=[2,8]/56=0,05(mol)`

`a)V_[H_2]=0,05.22,4=1,12(l)`

`b)V_[dd H_2 SO_4]=[0,05]/2=0,025(l)`

`c)m_[FeSO_4]=0,05.152=7,6(g)`

a) \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

b) \(2Ca+O_2\xrightarrow[]{t^o}2CaO\)

c) \(2Al+6HCl\rightarrow2AlCl_3\)

d) \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

e) \(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)

Fe+2HCl->FeCl₂+H₂

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

n _Fe=\(\dfrac{2,8}{56}=0,05mol\)

V_HCl=\(\dfrac{0,1}{2}=0,05l\)

=>V_H2=0,1.22,4=2,24l

`a) n_{Fe} = (2,8)/(56) = 0,05 (mol)`

`Fe + 2HCl -> FeCl_2 + H_2`

`b)` Theo PT: `n_{HCl} = 2n_{Fe} = 0,1 (mol)`

`=> V_{ddHCl} = (0,1)/2 = 0,05 (l)`

`c)` Theo PT: `n_{H_2} = n_{Fe} = 0,05 (mol)`

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

a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

`=> 56x + 24y = 8 (1)`

Ta có: `n_{H_2} = (4,48)/(22,4) = 0,2 (mol)`

PTHH: `Fe + H_2SO_4 -> FeSO_4 + H_2`

            `Mg + H_2SO_4 -> MgSO_4 + H_2`

Theo PT: `n_{H_2} = n_{Fe} + n_{Mg}`

`=> x + y = 0,2 (2)`

Từ `(1), (2) => x = y = 0,1`

`=>` \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{8}.100\%=70\%\\\%m_{Mg}=100\%-70\%=30\%\end{matrix}\right.\)

b) Theo PT: `n_{H_2SO_4} = n_{H_2} = 0,2 (mol)`

`=>` \(m_{ddH_2SO_4}=\dfrac{0,2.98}{49\%}=40\left(g\right)\)

c) Theo PT:

`n_{FeSO_4} = n_{Fe} = 0,1 (mol)`

`n_{MgSO_4} = n_{Mg} = 0,1 (mol)`

`=> m_{muối} = 0,1.152 + 0,1.120 = 27,2 (g)`

Fe+2HCl->FeCl₂+H₂

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

n _Fe=0,1 mol

=>m _FeCl2 =0,1.127 =12.7g

V_H2=0,1.22,4=2,24l

CM_HCl=\(\dfrac{0,2}{0,25}=0,8M\)