a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,75\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,75.65=48,75\left(g\right)\)
a) PTHH: Zn + 2HCl -> ZnCl\(_2\) + H\(_2\)
b)Ta có: n\(_{H_{ }2}\)=\(\dfrac{16.8}{22.4}\)=0.75 mol
theo PTHH => n\(_{ }h2\)=n\(_{Zn}\)=0.75 mol
=>m\(_{Zn}\)= 0.75x65 = 48.75g