n\(_{H_2}\) = \(\dfrac{2,24}{22,4}\)= 0,1 (mol)
a)
PTHH: Zn + 2HCl ----> ZnCl2 + H2\(\uparrow\)
mol: __0,1<-0,2<------------------0,1
b) m\(_{Zn}\)= 0,1 . 65 = 6,5 (g)
c)
PTHH: NaOH + HCl ----> NaCl + H2O
mol:__0,2<------0,2
m\(_{NaOH}\)= 0,2 . 40 = 8 (g)
mdd\(_{NaOH}\) = \(\dfrac{8.100}{15}\)= 53,33 (g)