\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{H_2}=n_{Fe}=\dfrac{10}{56}=\dfrac{5}{28}\left(mol\right)\\ \Rightarrow V_{H_2}=\dfrac{5}{28}.22,4=4\left(l\right)\)

`Fe + 2HCl -> FeCl_2 + H_2`

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

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

`a)m_[FeCl_2]=0,1.127=12,7(g)`

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

bạn từng câu lên sẽ dễ nhìn hơn 

M la j z bn?

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

Nếu có thể thì lần sau bạn nên đăng tách từng bài ra nhé!

Bài 1:

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) , ta được Mg dư.

Theo PT: \(n_{Mg\left(pư\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

\(\Rightarrow n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\)

\(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Bài 2:

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,15}{3}\) , ta được Al dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,05\left(mol\right)\\n_{H_2}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

Bài 3:

PT: \(2M+6HCl\rightarrow2MCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{4,704}{22,4}=0,21\left(mol\right)\)

Theo PT: \(n_M=\dfrac{2}{3}n_{H_2}=0,14\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{3,78}{0,14}=27\left(g/mol\right)\)

Vậy: M là nhôm (Al).

Bài 4:

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,2\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}>\dfrac{0,2}{5}\) , ta được P dư.

Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,08\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,08.142=11,36\left(g\right)\)

Bạn tham khảo nhé!

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

___0,3____0,6_____0,3____0,3 (mol)

a, \(m_{Zn}=0,3.65=19,5\left(g\right)\)

b, \(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

c, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

Bạn tham khảo nhé!

Zn+2HCl->ZnCl₂+H₂

0,3---0,6-----0,3----0,3 mol

=>m _Zn=0,3.65=19,5g

=>m _HCl=0,6.35,6=21,9g

=>m _ZnCl2=0,3.136=40,8g