\(H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\)

0,2        0,2            0,2       0,2 

\(b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

\(m_{CuO}=0,2.80=16\left(g\right)\)

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(m_{H_2O}=0,2.18=3,6\left(g\right)\)

Ta có: \(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT ta có: \(n_{CuO}=n_{Cu}=0,025\left(mol\right)\)
Khối lượng đồng (II) oxit phản ứng là:
\(m_{CuO}=0,025.80=2\left(g\right)\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

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

Theo PTHH: \(n_{CuO}:n_{Cu}=1:1\Rightarrow n_{CuO}:n_{Cu}=0,025\left(mol\right)\)

\(\Rightarrow m_{CuO}=n_{CuO}.M_{CuO}=0,025.80=2\left(g\right)\)

PTHH. CuO + H2 -> H2O + Cu

Theo bài:
nCu = \(\dfrac{1,6}{64}=0,025\left(mol\right)\)

Theo pthh và bài có:

nCuO = nCu = 0,025 (mol)

=>mCuO = 0,025 . 80 = 2 (g)

Ta có: \(n_{Cu}=\dfrac{0,32}{64}=0,005\left(mol\right)\)

a, PTHH: CuO + H₂ -t^o-> H₂ + CuO

b, Ta có: \(n_{CuO}=n_{H_2}=n_{Cu}=0,005\left(mol\right)\)

=> \(m_{CuO}=0,005.80=0,4\left(g\right)\)

c) \(V_{H_2\left(đktc\right)}=0,005.22,4=0,112\left(l\right)\)

n_Cu=m/M=0,32/64=0,005 (mol)

PT:

CuO + H₂ -t⁰-> Cu +H₂O

1...........1............1...........1 (mol)

0,005<- 0,005<-0,005->0,005 (mol)

b) => m_CuO=n.M=0,005.80=0,4 (g)

c) V_H2=n.22,4=0,005.22,4=0,112(lít)

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

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

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,1\cdot64=6,4\left(g\right)\\m_{CuO}=80\cdot0,1=8\left(g\right)\end{matrix}\right.\)

d) Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

\(\Rightarrow\) CuO còn dư, Hidro p/ứ hết

\(\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\) \(\Rightarrow m_{CuO\left(dư\right)}=80\cdot0,05=4\left(g\right)\)