Question

Answer 1

Câu1

Fe+2HCl->FeCl₂+H₂

0,12-0,24-0,12-----0,12 mol

n _Fe=\(\dfrac{6,72}{56}\)=0,12 mol

->V_H2=0,12.22,4=2,688l

->m _FeCl2=0,12.127=15,24g

->V_HCl=\(\dfrac{0,24}{0,5}\)=0,48l=480ml

Answer 2

bài3

MnO₂+4HCl->MnCl₂+Cl₂+2H₂O

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

m MnO2=\(\dfrac{104,4}{87}=1,2mol\)

=>V_Cl2=1,2.22,4=26,88l

Answer 3

Câu 2.

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)

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

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

x          3x           x            1,5x

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

 y         2y           y           y

\(\Rightarrow\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11}\cdot100\%=49,1\%\)

\(\%m_{Fe}=100\%-49,1\%=50,9\%\)

b)\(\Sigma n_{HCl}=3x+2y=3\cdot0,2+2\cdot0,1=0,8mol\)

\(V_{HCl}=\dfrac{n}{C}=\dfrac{0,8}{0,5}=1,6M\)

Answer 4

Câu 1.

\(n_{Fe}=\dfrac{6,72}{56}=0,12mol\)

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

0,2       0,4          0,2       0,2

\(V=0,2\cdot22,4=4,48l\)

\(m=0,2\cdot127=25,4g\)

\(V_{HCl}=\dfrac{n}{C}=\dfrac{0,4}{0,5}=0,8M\)

Answer 5

bài 5

ZnO+2Hcl->ZnCl2+H2O

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

Zn+2HCl->ZnCl2+H2

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

n H2=\(\dfrac{2,24}{22,4}\)=0,1 mol

=>m Zn=0,1.65=6,5g

=>m ZnO=4,05g

->n ZnO=\(\dfrac{4,05}{81}\)=0,05 mol

=>%mZn=\(\dfrac{6,5}{10,55}.100\)=61,61%

=>%mZnO=38,29%

=>m dd HCl=109,5g

Answer 6

bài 4

a)M+2HCl->MCl₂+H₂

n _H2=\(\dfrac{2,8}{22,4}\)=0,125 mol

->\(\dfrac{3}{M_M}=\)0,125

=>M_M=24 đvC

=>M là Mg(magie)

b) đặt M hóa trị 1

2M+2HCl->2MCl+H₂

n H2=\(\dfrac{3,864}{22,4}\)=0,1725 mol

=>\(\dfrac{3,105}{M}=\)0,345

=>M =9 đvC

ta có bảng:

n     1      2         3

M    9    18       27

=>n =3 =>M =27 nhận 

=>M là nhôm (Al)