oxit: \(Fe_2O_3\) (sắt III oxit)

axit:

\(H_2SO_4\) (axit sunfuric)

\(H_2S\) (axit sunfua)

bazơ:

\(Fe\left(OH\right)_2\) (sắt II hidroxit)

muối:

\(ZnCl_2\) (kẽm clorua)

\(FeSO_4\) (sắt II sunfat)

cái H2S đọc tên sai

\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)

tỉ lệ:          3     :   2      :      1

n(mol)     0,1--->`1/15` -->`1/30`

\(m_{Fe_3O_4}=n\cdot M=\dfrac{1}{30}\cdot\left(56\cdot3+16\cdot4\right)=\approx7,73\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=\dfrac{1}{15}\cdot22,4=\approx1,5\left(l\right)\)

người ta thu được sản phẩm gì ạ: FeO,Fe2O3 hay Fe3O4 ?

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{19,6}{2+32+16\cdot4}=0,2\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ:        1     :     1        :      1        :      1

n(mol)     0,2<---0,2------>0,2-------->0,2

\(m_{ZnSO_4}=n\cdot M=0,2\cdot\left(65+32+16\cdot4\right)=32,2\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

Đặt \(\left\{{}\begin{matrix}n_{CaCO_3}=x\\n_{MgCO_3}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_{hh}=100x+84y=14,2\left(g\right)\)   (1)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\)

    x                      x          x        ( mol )

\(MgCO_3\rightarrow\left(t^o\right)MgO+CO_2\)

    y                       y            y         ( mol )

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                        0,15     0,15                 ( mol )

\(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)

\(\Rightarrow n_{CO_2}=x+y=0,15\left(mol\right)\)          (2)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,1.100}{14,2}.100=70,42\%\\\%m_{MgCO_3}=100-70,42=29,58\%\end{matrix}\right.\)

\(m_{rắn}=m_{CaO}+m_{MgO}=0,1.56+0,05.40=7,6\left(g\right)\)

\(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)

 \(n_{CO_2}=\dfrac{m+28}{44}\left(mol\right)\)

 \(n_{H_2O}=\dfrac{m+2}{18}\left(mol\right)\)

Ta có: \(n_{X\left(ankin\right)}=n_{CO_2}-n_{H_2O}\)

\(\Leftrightarrow0,02=\dfrac{m+28}{44}-\dfrac{m+2}{18}\)

\(\Leftrightarrow7,92=9\left(m+28\right)-22\left(m+2\right)\)

\(\Leftrightarrow7,92=9m+252-22m-44\)

\(\Leftrightarrow-13m=-200,08\)

\(\Leftrightarrow m=\dfrac{5002}{325}\left(g\right)\)

em cảm ơn ạ

\(H_2+Ag_2O\rightarrow\left(t^o\right)2Ag+H_2O\\ n_{Ag}=\dfrac{10,7}{108}\)

Em ơi 10,7/108?? Em ơi coi lại đề kq sẽ ra bị xấu

Đề đúng rồi ạ

\(C+O_2\rightarrow\left(t^o\right)CO_2\\ n_C=\dfrac{96}{12}=8\left(mol\right)=n_{O_2}\\ V_{kk}=V_{O_2\left(đktc\right)}.5=\left(22,4.8\right).5=896\left(l\right)\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\\ n_{O_2}=2.n_{CH_4}=2.0,2=0,4\left(mol\right)\\ a,V_{kk}=5.V_{O_2\left(đktc\right)}=5.\left(0,4.22,4\right)=44,8\left(l\right)\\ b,m_{CO_2}=0,2.44=8,8\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{40}{56\cdot2+16\cdot3}=0,25\left(mol\right)\\ PTHH:Fe_2O_3+3H_2-^{t^o}>2Fe+3H_2O\)

n(mol)       0,25->0,75-------->0,5---->0,75

\(m_{Fe}=n\cdot M=0,5\cdot56=28\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,75\cdot22,4=16,8\left(g\right)\)