\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ \left(mol\right)...0,2......\leftarrow...............0,2\\ m_{Zn}=0,2.65=13\left(g\right)\)
PTHH: Zn + 2HCl _____> ZnCl2 + H2 (1)
Ta có: theo (1): n\(H_2\)(đktc)=\(\dfrac{4.48}{22.4}\)=0.2 (mol)
theo (1): nZn = n\(H_2\)= 0.2(mol)
=> mZn = 0.2 . 65 = 13(g)
Vậy giá trị m bằng 13(g)