Question

Answer 1

a) \(\left\{{}\begin{matrix}n_{CO_2}+n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\\overline{M}=\dfrac{44.n_{CO_2}+2.n_{H_2}}{n_{CO_2}+n_{H_2}}=15.2=30\left(g/mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{H_2}=0,05\left(mol\right)\end{matrix}\right.\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

          0,05<---0,1<------------0,05

            \(FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\)

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

=> \(\left\{{}\begin{matrix}m_{Fe}=0,05.56=2,8\left(g\right)\\m_{FeCO_3}=0,1.116=11,6\left(g\right)\\m_{FeO}=21,6-2,8-11,6=7,2\left(g\right)\end{matrix}\right.\)

b) \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

PTHH: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

              0,1---->0,2

=> n_HCl = 0,1 + 0,2 + 0,2 = 0,5 (mol)

=> \(V_{dd.HCl}=\dfrac{0,5}{0,5}=1\left(l\right)\)

Answer 2

a)

PTHH:

\(Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\\ FeO+2HCl\rightarrow FeCl_2+H_2O\left(2\right)\\ FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\left(3\right)\)

hh khí thu được: H₂, CO₂

\(\overline{M}_{hh}=15.2=30\left(g/mol\right)\)

\(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Áp dụng sơ đồ đường chéo, ta có:

\(\dfrac{n_{H_2}}{n_{CO_2}}=\dfrac{44-30}{30-2}=\dfrac{1}{2}\)

=> \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{1+2}.0,15=0,05\left(mol\right)\\n_{CO_2}=0,15-0,05=0,1\left(mol\right)\end{matrix}\right.\)

Theo (1): \(n_{Fe}=n_{H_2}=0,05\left(mol\right)\)

=> m_Fe = 0,05.56 = 2,8 (g)

Theo (2): \(n_{FeCO_3}=n_{CO_2}=0,1\left(mol\right)\)

=> m_FeCO3 = 0,1.116 = 11,6 (g)

=> m_FeO = 21,6 - 2,8 - 11,6 = 7,2 (g)

b) \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo (1), (2), (3): \(n_{HCl}=2n_{Fe}+2n_{FeO}+2n_{FeCO_3}=0,5\left(mol\right)\)

=> \(V_{dd.HCl}=\dfrac{0,5}{0,5}=1\left(l\right)\)

Answer 3

Gọi \(\left\{{}\begin{matrix}n_{H_2}=x\\n_{CO_2}=y\end{matrix}\right.\) ( mol )

\(n_{hhk}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

`x`      `2x`                       `x`      ( mol )

\(FeO+2HCl\rightarrow FeCl_2+H_2O\) (2)

\(FeCO_3+2HCl\rightarrow FeCl_2+CO_2\uparrow+H_2O\)

  `y`           `2y`                             `y`                 ( mol )

\(\dfrac{M_{hhk}}{M_{H_2}}=15\)

\(\rightarrow\dfrac{2x+44y}{x+y}=15.2\)

\(\Leftrightarrow2x+44y=30x+30y\)  

\(\Leftrightarrow28x-14y=0\) (1)

\(\rightarrow n_{hhk}=x+y=0,15\) (2)

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

\(\rightarrow m_{Fe}=0,05.56=2,8\left(g\right)\)

\(m_{FeCO_3}=0,1.116=11,6\left(g\right)\)

\(\rightarrow m_{FeO}=21,6-2,8-11,6=7,2\left(g\right)\)

\(\left(2\right)\rightarrow n_{HCl}=2.\dfrac{7,2}{72}=0,2\left(mol\right)\)

\(V_{HCl}=\dfrac{0,05.2+0,1.2+0,2}{0,5}=1\left(l\right)\)