\(V_{H_2}=0,6.24,79=14,874(l)\)

\(VH2=0,6.22,4=13,44\)\((l)\)

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

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

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`

\(n_{H_2}=\dfrac{3}{2}=1,5mol\)

\(V_{O_2}=2.22,4=44,8l\)

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

\(V_{N_2}=0,1.22,4=2,24\left(l\right)\)

\(V_{CO_2}=0,45.22,4=10,08\left(l\right)\)

\(V_{hh}=5,6+2,24+10,08=17,92\left(l\right)\)

$V_{hh}=22,4(0,25+0,1+0,45)=17,92(lít)$

Sửa đề: 0,5 mol → 0,5 M

a, Ta có: \(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{HCl}=2n_{Zn}=0,08\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,08}{0,5}=0,16\left(l\right)\)

b, Theo PT: \(n_{H_2}=n_{Zn}=0,04\left(mol\right)\Rightarrow V_{H_2}=0,04.22,4=0,896\left(l\right)\)

Sửa đề 0,5mol ⇒ 0,5M

\(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

      0,04    0,08                      0,04

a) \(V_{ddHCl}=\dfrac{0,08}{0,5}=0,16\left(l\right)\)

b) \(V_{H2\left(dktc\right)}=0,04.22,4=0,896\left(l\right)\)

 Chúc bạn học tốt

a)zn+2hcl→zncl2 +h2

1mol  2mol    1mol     1mol

0,15mol  0,3mol             0,15mol

nzn=\(\frac{9,75}{65}\)=0,15mol 

Vh2 =0,15 .22,4=3,36 (lít)

Chcl =\(\frac{nhcl}{vhcl}\)

         =\(\frac{0,3}{0,3}\)= 1( đổi 300ml = 0,3 lít )

Không có đồng 3 oxit nhá bạn

\(n_{Zn}=\dfrac{130}{65}=2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=2mol\\ V_{H_2}=2.22,4=44,8l\\ 1000ml=1l\\ n_{HCl}=2.2=4mol\\ C_{M_{HCl}}=\dfrac{4}{1}=4M\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{130}{65}=2mol\) 

PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

TL;       1        2          1          1

mol:      2 \(\rightarrow\) 4

\(m_{HCl}=n.M=4.36,5=146g\) 

đổi 1000 ml= 1l

\(C\%_{ddHCl}=\dfrac{m_{HCl}}{V_{HCl}}.100\%=\dfrac{146}{1}.100=14600\%\)

số hơi lớn em xem lại đề nhé