Đề cho bao nhiêu gam axit clohidric bạn nhỉ?
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{3}{36,5}=\dfrac{6}{73}\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{\dfrac{6}{73}}{2}\), ta được Zn dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{3}{73}\left(mol\right)\)
\(\Rightarrow V_{H_2}=\dfrac{3}{73}.22,4=\dfrac{336}{365}\left(l\right)\)