Hóa học

\(\%O=100\%-20,72\%-28,82\%=50,46\%\)

Giả sử: A có \(100\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=20,72\left(g\right)\\m_S=28,82\left(g\right)\\m_O=50,46\left(g\right)\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}n_{Na}=0,9\left(mol\right)\\n_S=0,9\left(mol\right)\\n_O=3,2\left(mol\right)\end{matrix}\right.\)

\(n_{Na}:n_S:n_O=0,9:0,9:3,2=2:2:7\)

`=>` CTHH A: \(Na_2S_2O_7\)

Đặt CTTQ: \(Na_xS_yO_z\)

\(x:y:z=\dfrac{20.72}{23}:\dfrac{28.82}{32}:\dfrac{\left(100-20.72-28.82\right)}{16}=0,90:0,90:3,15=1:1:3,5\left(không.thoả.mãn\right)\)

Vậy không tồn tại CTPT thoả mãn đề.

\(2M\left(OH\right)_n\rightarrow\left(t^o\right)M_2O_n+nH_2O\)

\(2MOH\rightarrow M_2O+H_2O\)

\(\Rightarrow n=1\)

\(n_{H_2SO_4}=\dfrac{100.9,8\%}{98}=0,1\left(mol\right)\)

`CuO+H_2SO_4->CuSO_4+H_2O`

  0,1                          0,1                      ( mol )

`m_{CuO}=0,1.80=8(g)`

`m_{CuSO_4}=0,1.160=16(g)`

CuO+H₂SO₄->CuSO₄+H₂O

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

m_H2SO4=9,8g

=>n_H2SO4=\(\dfrac{9,8}{98}\)0,1 mol

=>m _CuO=0,1.80=8g

=>m _CuSO4=0,1.160=16g

\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\) ; \(n_{AgNO_3}=\dfrac{400.68\%}{170}=1,6\left(mol\right)\)

\(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\downarrow\)

 0,1            < 1,6                                         ( mol )

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

\(m_{ddspứ}=200+400-0,2.143,5=571,3\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{Ba\left(NO_3\right)_2}=\dfrac{0,1.261}{571,3}.100=4,56\%\\C\%_{AgNO_3\left(dư\right)}=\dfrac{\left(1,6-0,2\right).170}{571,3}.100=41,65\%\end{matrix}\right.\)

Quy hỗn hợp thành \(\left\{{}\begin{matrix}FeO\\Fe_2O_3\end{matrix}\right.\)

Ta có: \(\dfrac{n_{FeO}}{n_{Fe_2O_3}}=\dfrac{1}{1}\) \(\Rightarrow n_{FeO}=n_{Fe_2O_3}=x\left(mol\right)\)

\(\Rightarrow m_{hh}=72x+160x=23,2\)

\(\Leftrightarrow x=0,1\left(mol\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

0,1          0,1                           ( mol )

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

   0,1             0,3                                  ( mol )

\(\Rightarrow a=n_{H_2SO_4}=0,1+0,3=0,4\left(mol\right)\)

tỉ lệ mol của Fe0 và Fe2O3 là 1:1( mình nhấn nhầm )

Gọi \(\left\{{}\begin{matrix}n_{FeO}=n_{Fe_2O_3}=x\left(mol\right)\\n_{Fe_3O_4}=y\left(mol\right)\end{matrix}\right.\)

`=> 72x + 160x + 232y = 23,2`

`=> x + y = 0,1`

PTHH: \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\) (1)

            \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\) (2)

            \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\) (3)

Theo PT: \(n_{H_2SO_4}=n_{FeO}+3n_{Fe_2O_3}+4n_{Fe_3O_4}=4a+4b\)

`=>` \(a=n_{H_2SO_4}=4.\left(a+b\right)=0,1.4=0,4\left(mol\right)\)

$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
$ZnO + H_2SO_4 \to ZnSO_4 + H_2O$
$MgO + H_2SO_4 \to MgSO_4 + H_2O$

Theo PTHH : $n_{H_2O} = n_{H_2SO_4} = 0,05(mol)$
Bảo toàn khối lượng : 

$m_{muối} = m_{hh} + m_{H_2SO_4} - m_{H_2O} = 2,81 + 0,05.98 - 0,05.18 = 6,81(gam)$

^{Fe2O3+ H2SO4--->FeSO4+H2O}

^{ZnO +H2SO4---> ZnSO4+ H2O}

^{MgO+ H2SO4---->MgSO4+H2O}

^{Do nH2SO4=0,05}

^{=>mH2SO4= 0,05.98=4,9 (g)}

^{Theo pt :nH2O=nH2SO4 =0,05 MOL}

^{mH2O=0,05.18=0,9 (g)}

^{ADDLBTKL ta có :2,81+4,9=m muối+0,9}

^{=> m muối=6,81(g)}

a)

$Na_2O + 2HCl \to 2NaCl + H_2O$
b)

$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$

Theo PTHH : 

$n_{NaCl} = 2n_{Na_2O} = 0,2(mol)$
$m_{NaCl} = 0,2.58,5 = 11,7(gam)$

c)

$n_{HCl} = 2n_{Na_2O} = 0,2(mol)$

$C_{M_{HCl}} = \dfrac{0,2}{0,3} = 0,67M$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$

Theo PTHH : $n_{H_2SO_4} = n_{H_2} = \dfrac{ \dfrac{5m}{67}}{2} = \dfrac{5m}{134}(mol)$

Bảo toàn khối lượng : 

$m_{hh} + m_{H_2SO_4} = m_{muối} + m_{H_2}$
$\Rightarrow m + \dfrac{5m}{134}.98 = 61,4 + \dfrac{5m}{67}$

$\Rightarrow m = 13,64(gam)$

tại sao chiếc thuyền to lại nổi mà cây kim lại chìm?

Vì sao ma sát đầu bút bi có thể hút giấy?

Vì sao có nước xun quanh thành cốc khi đá tan?

Gọi \(m_{KClO_3}=m_{KMnO_4}=m_{MnO_2}=a\left(mol\right)\)

`=>` \(\left\{{}\begin{matrix}n_{KClO_3}=\dfrac{a}{122,5}\left(mol\right)\\n_{KMnO_4}=\dfrac{a}{158}\left(mol\right)\\n_{MnO_2}=\dfrac{a}{87}\left(mol\right)\end{matrix}\right.\)

PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)

              \(\dfrac{a}{122,5}\)-------------------->\(\dfrac{3a}{245}\)

           \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

               \(\dfrac{a}{158}\)--------------------------------->\(\dfrac{a}{316}\)

           \(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)

           \(\dfrac{a}{122,5}\)------------------------>\(\dfrac{6a}{245}\)

            \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

               \(\dfrac{a}{158}\)---------------------------------------------->\(\dfrac{5a}{316}\)       

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

               \(\dfrac{a}{87}\)------------------------------>\(\dfrac{a}{87}\)

`=>` \(\left\{{}\begin{matrix}\dfrac{3a}{245}>\dfrac{a}{316}\\\dfrac{6a}{245}>\dfrac{5a}{316}>\dfrac{a}{87}\end{matrix}\right.\)

`=> KClO_3` đều cho nhiều `Cl_2` và `O_2` hơn