- ta có :
7,3% = \(\dfrac{m_{HCl}}{50}\).100%
\(\Rightarrow\) mHCl = 3,65 g
\(\Rightarrow\)nHCl = \(\dfrac{3,65}{36,5}\) = 0,1 mol
a) Zn + 2HCl \(\rightarrow\)ZnCl2 + H2 \(\uparrow\)
0,05..\(\leftarrow\)0,1................\(\rightarrow\)0,05
b) m = 0,05 . 65 = 3,25 g
VH2 = 0,05 . 22,4 = 1,12 (l)