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

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

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

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

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

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

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

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

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

Chúc em học tốt !!

\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
pthh: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
        0,1                 0,1       0,1
=>\(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{H_2O}=0,1.18=1,8\left(g\right)\end{matrix}\right.\)

a) Phản ứng

CuO   +   H 2   → t o   Cu   +   H 2 O (1)

(mol) 0,3          0,3 ← 0,3

b) Ta có: n Cu = 19,2/64 = 0,3 (mol)

Từ (1) →  n Cu  = 0,3 (mol) → m CuO = 0,3 x 80 = 24 (gam)

Và n H 2 = 0,3 (mol) → V H 2 =0,3 x 22,4 = 6,72 (lít)

a)PTHH: CuO+H2=>Cu+H2O

 nCu=0,05g

PTHH: CuO+H2=>Cu+H2O

               0,05<-0,05<-0,05->0,05

=> VH2 =0,05.22,4=1,12 lit

nO2=2,2422,4=0,1(mol)nO2=2,2422,4=0,1(mol)

PTHH: 2Cu + O₂ --t^o--> 2CuO

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

=> m_CuO = 0,2.80 = 16(g)

a)PTHH: CuO+H2=>Cu+H2O

 nCu=0,05g

PTHH: CuO+H2=>Cu+H2O

               0,05<-0,05<-0,05->0,05

=> VH2 =0,05.22,4=1,12 lit

b) mH20=18.0,05=0,9 gam

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(n_{CuO}=\dfrac{40}{80}=0.5\left(mol\right)\)

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

\(1...........1\)

\(0.5............0.1\)

\(LTL:\dfrac{0.5}{1}>\dfrac{0.1}{1}\Rightarrow CuOdư\)

\(m_{cr}=m_{CuO\left(dư\right)}+m_{Cu}=\left(0.5-0.1\right)\cdot80+0.1\cdot64=39.4\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,1}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\)

⇒ n_{CuO dư} = 0,4 (mol)

⇒ m _{chất rắn} = m_Cu + m_{CuO dư} = 0,1.64 + 0,4.80 = 38,4 (g)

Bạn tham khảo nhé!

`PTHH:`

  `CuO + H_2 -t^o-> Cu + H_2 O`

`0,045`   `0,045`       `0,045`             `(mol)`

`n_[CuO]=[3,6]/80=0,045(mol)`

`a)m_[Cu]=0,045.64=2,88(g)`

`b)V_[H_2]=0,045.22,4=1,008(l)`