Câu 1:
\(C\%_{CaCl_2}=\dfrac{20}{250+20}.100\%=7,41\%\)
Câu 2:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1-->0,2-------->0,1---->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
\(b,V_{dd}=\dfrac{0,2}{2}=0,1\left(l\right)\\ \rightarrow C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,1}=1M\)