a. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b. \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PTHH: \(n_{Zn}=n_{H_2}=0,4\left(mol\right)\)
\(a=m_{Zn}=0,4\cdot65=26\left(g\right)\)
a, `Zn + 2HCl -> ZnCl_2 + H_2`
`b, n_(H_2) = (8,96)/(22,4) = 0,4 mol`
`=> n_(Zn) = 0,4 mol`
`=> m_(Zn) = 0,4 xx 65 = 26 g`.