Question

trời ơi cứu tuii trờiii oii=))sos

Answer 1

4. a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PTHH: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

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

c) \(C_M=\dfrac{n_{ct}}{V_{HCl}}=\dfrac{0,1}{0,8}=0,125M\)

d) \(H_2+CuO\rightarrow Cu+H_2O\)

\(n_{CuO}=\dfrac{0,8}{80}=0,1\left(mol\right)\)

Ta có: \(\dfrac{n_{H_2}}{1}=\dfrac{n_{CuO}}{1}\)

Vậy không có chất nào dư

\(\Rightarrow n_{Cu}=n_{H_2}=0,1\left(mol\right)\)

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