Các PTHH:

$2K+2H_2O\to 2KOH+H_2$

$Ca+2H_2O\to Ca(OH)_2+H_2$

`n_{H_2}=(3,7185)/(24,79)=0,15\ (mol)`

Theo phương trình:

`n_{OH^-}=2n_{H_2}=0,3\ (mol)`

Để trung hoà 1/2 dung dịch X

`\to n_{H^+}=n_{HCl}=1/2n_{OH^-}=0,15\ (mol)`

`\to V_{HCl}=(0,15)/2=0,075\ (l)=75\ (ml)`