PTHH :Zn + 2HCl \(\rightarrow\) ZnCl\(_2\) + H\(_2\)
n\(_{Zn}\)= \(\dfrac{0,5}{65}\)=0,008 mol
Từ pt ta có n\(_{H_2}\)=n\(_{Zn}\)=0,008 mol
V\(_{H_2}\)=0,008 . 22,4 = 3,57 l
PTHH : 3H\(_2\)+ Cu\(_2\)O\(_3\)\(\rightarrow\)2Cu +3H\(_2\)O
n\(_{H_2}\)=0,008
n\(_{Cu_2O_3}\)= 0,034
\(\dfrac{0,008}{3}< \dfrac{0,034}{1}\) VẬY Cu\(_2\)O\(_3\) dư
n\(_{Cu}\)=\(\dfrac{2}{3}\)n\(_{H_2}\)=0,005 mol
m\(_{Cu}\)=0,005.64=0,32