\(Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = n_{ZnSO_4} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ m_{Zn} = 0,3.65 = 19,5(gam)\\ m_{ZnSO_4} = 0,3.161 = 48,3(gam)\)
a) Zn+H2SO4→ZnSO4+H2
b) Ta có : VH2= 6,72(l) → nH2= nZn=nZnSO4=\(\dfrac{6,72}{22,4}\)=0,3(mol)
mZn=0,3.65=19,5(gam)
mZnSO4=0,3.161=48,3(gam)