\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)

\(\Rightarrow H=\dfrac{\dfrac{116}{15}}{11,6}.100\%\approx66,67\%\)

\(n_{Fe_3O_4}=\dfrac{1}{30}\left(mol\right)\) chứ không phải \(\dfrac{1}{3}\) em nhé.

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pt: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2mol\)

\(n_{Fe}=\dfrac{6,72}{56}=0,12mol\\ 2Fe+O_2\xrightarrow[]{t^0}2FeO\)

0,12        0,06       0,12

\(m_{FeO}=0,12.72=8,64g\\ V_{O_2}=0,06.24,79=1,4874l\)

fgsr

4Fe\(_{ }\) + 3O2 -> 2Fe2O3

a, \(2MgO+O_2\underrightarrow{t^o}2MgO\)

b, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

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

d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(a,2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

\(b,4Al+3O_2\rightarrow2Al_2O_3\)

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

\(d,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Bài 1 :

\(m_{NaCl}=\dfrac{1000.0.9}{100}=9\left(g\right)\)

\(m_{dm}=1000-9=991\left(g\right)\)

Vậy cần pha 9g muối khan Nacl vào 991 g nước để cho được 1000g dd NaCl 0,9%

Bài 2 :

\(a,m_{dd}=m_{ct}+m_{dm}=1600+40=1640\left(g\right)\)

\(C\%_{Na_2CO_3}=\dfrac{40}{1640}.100\%\simeq2,44\%\)

\(b,m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{29,4}{250}.100\%=11,76\%\)

Gọi CTTQ là : XO₃ 

\(a,\rightarrow M_A=80\)

\(\rightarrow M_X=80-\left(16.3\right)=32\left(\dfrac{g}{mol}\right)\)

Vậy X là S ( lưu huỳnh )

\(\%m_{S\left(SO_3\right)}=\dfrac{32}{80}.100\%=40\%\)

Gọi CTHH cần tìm là N_xH_y.

\(\Rightarrow x:y=\dfrac{82,35}{14}:\dfrac{17,65}{1}\approx1:3\)

→ CTHH cần tìm có dạng (NH₃)_n

Mà: \(d_{N_xH_y/H_2}=8,5\Rightarrow M_{N_xH_y}=8,5.2=17\left(g/mol\right)\)

\(\Rightarrow n=\dfrac{17}{14+1.3}=1\)

Vậy: CTHH cần tìm là NH₃.