... câu a) có mỗi số liệu của Fe thì sao tính chất dư được :v

$n_{Zn} = \dfrac{6,5}{65} = 0,1(mol) \\ PTHH: Zn + 2HCl \to ZnCl_2 + H_2 \\$$n_{H_2} = n_{Zn} = 0,1(Mol) \\ V_{H_2} = 0,1.22,4 = 2,24l \\b) PTHH: H_2 + CuO \xrightarrow[]{t^o} Cu + H_2O \\ n_{CuO} = \dfrac{12}{64} = 0,15(mol) \\ \to CuO dư$ $\\ n_{H_2} = n_{Cu} = 0,1(mol \\ m_{Cu} = 0,1.64 = 6,4(gam)$

Zn+2HCl->ZnCl2+H2

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

n Zn=0,1 mol

=>VH2=0,1.22,4=2,24l

H2+CuO-to>Cu+H2O

          0,15-----0,15

n CuO=0,15 mol

=>H2 dư

=>m Cu=0,15.64=9,6g

\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)

\(n_{CuO}=\dfrac{36}{80}=0.45\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.3.....................................0.3\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

\(0.3.......0.3.....0.3....0.3\)

\(m_{Cr}=m_{CuO\left(dư\right)}+m_{Cu}=\left(0.45-0.3\right)\cdot80+0.3\cdot64=31.2\left(g\right)\)

\(m_{H_2O}=0.3\cdot18=5.4\left(g\right)\)

Chúc em học tốt !!

Zn+H₂SO₄→ZnSO₄+H₂ bạn biến đổi nó ra phương trình này kiểu gì vậy?

`Fe+2HCl->FeCl_2+H_2`

0,15--0,3---0,15----0,15 mol

`n_(Fe)=(8,4)/56=0,15 mol`

`->V_(H_2)=0,15.22,4=3,36l`

c) `CuO+H_2->Cu+H_2O`(to)              

             0,15---0,15 mol

`n_(CuO)=16/80=0,2 mol`

=>CuO dư

`->m_(Cu)=0,15.64=9,6g`

a. \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b. \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)

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

0,5-------1---------0,5-----0,5

Theo PTHH: \(\Rightarrow n_{H_2}=n_{Fe}=0,5\left(mol\right)\)

\(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)

c. \(H_2+CuO\rightarrow Cu+H_2O\)

  0,5-------0,5-----0,5----0,5

\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,5.64=32\left(g\right)\)

tóm tắt ạ

\(n_{HCl}=0.2\cdot1=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(........0.2..............0.1\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(0.1.......0.1....0.1\)

\(\Rightarrow CuOdư\)

\(m_{Cu}=0.1\cdot64=6.4\left(g\right)\)

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

`n_[HCl]=[10,95]/[36,5]=0,3(mol)`

`Fe + 2HCl -> FeCl_2 + H_2`

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

Ta có:`[0,1]/1 < [0,3]/2`

    `=>HCl` dư

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

`b)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

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

`=>m_[Cu]=0,1.64=6,4(g)`

\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

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

bđ        0,1     0,3

pư        0,1     0,2 

spư       0        0,1         0,1        0,1

\(\rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, PTHH: \(H_2+CuO\xrightarrow[]{t^o}Cu+H_2O\)

                0,1------------>0,1

\(\rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

a) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)

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

0,5-------1---------0,5------0,5

b) \(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)

c) \(H_2+CuO\rightarrow Cu+H_2O\)

  0,5-----0,5------0,5----0,5

Khối lượng đồng tạo thành: \(m_{Cu}=n_{Cu}.64=0,5.64=32\left(g\right)\)

THIẾU tóm Tắt

a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

PTHH: `Fe + 2HCl -> FeCl_2 + H_2`

         0,5-------------------------->0,5`

b) `V_{H_2} = 0,5.22,4 = 11,2 (l)`

c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

                          0,5---->0,5

`=> m_{Cu} = 0,5.64 = 32 (g)`

\(Fe+2HCl\underrightarrow{t^o}FeCl_2+H_2\)

\(1mol\)                         \(1mol\)

\(0,5mol\)                    \(0,5mol\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)

\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)

\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

\(1mol\)            \(1mol\)

\(0,5mol\)       \(0,5mol\)

\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)