\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\\ PTHH:Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\\ \left(mol\right)..0,5...\rightarrow1............0,5.........0,5\\a, V_{H_2}=0,5.22,4=11,2\left(l\right)\\ b,m_{CaCl_2}=0,5.111=55,5\left(g\right)\\ c,C_{M_{HCl}}=\dfrac{1}{0,5}=2\left(M\right)\)