Câu 15 : 

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

     0,4----->1,2------->0,4------>0,6

\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{43.8.100\%}{25\%}=175,2\left(g\right)\)

\(m_{ddspu}=10,8+175,2-0,6.2=184,8\left(g\right)\)

\(C\%_{AlCl3}=\dfrac{0,4.133,5}{184,8}.100\%=28,9\%\)

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

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, Ta có: m _{dd sau pư} = 8,1 + 200 - 0,45.2 = 207,2 (g)

Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,35_____0,7___________0,35 (mol)

a, \(m_{Zn}=0,35.65=22,75\left(g\right)\)

b, \(C\%_{HCl}=\dfrac{0,7.36,5}{200}.100\%=12,775\%\)

\(a)n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,35        0,7            0,35         0,35

\(m_{Zn}=0,35.65=22,75g\\ b)C_{\%HCl}=\dfrac{0,7.36,5}{200}\cdot100\%=12,775\%\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Zn}=0,2\left(mol\right)\\n_{HCl}=2n_{Zn}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

a)

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

b)

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

Theo PT:\(n_{HCl}=2n_{Fe}=0,2mol\)

\(\Rightarrow C_MHCl=\dfrac{0,2}{0,2}=1M\)

c)

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

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

nNa2O= 12,4/62=0,2(mol)

a) PTHH: Na2O + H2O -> 2 NaOH

0,2_________0,2____0,4(mol)

b) VddNaOH=2(l)

=>CMddNaOH=0,4/0,2=2(M)

Chúc em học tốt!

CuO + 2Hcl = CuCl2 + H20

0.075...0,15...0,075

Cm = n/V = 0,15/0,4= 0,375

mCuCl2 = 0,075.135=10,125

a) CuO + 2HCl → CuCl₂ + H₂O

b)0,075.....0,15.......0,075

⇒C_M\(_{HCl}\)=\(\dfrac{1,15}{0,4}\)=0,375M

c)m\(_{CuCl_2}\)=0.075.135=10,125(g)

`a)PTHH:`

`Mg + 2HCl -> MgCl_2 + H_2`

`0,3`      `0,6`             `0,3`       `0,3`    `(mol)`

`b) n_[Mg] = [ 7,2 ] / 24 = 0,3 (mol)`

 `=> V_[H_2] = 0,2 . 22,4 =6,72 (l)`

`c) C_[M_[HCl]] = [ 0,6 ] / [ 0,3 ] = 2 (M)`