Đáp án:
a) Đổi $100 ml = 0,1 l$
PTHH :
$CO_2 + Ca(OH)_2 \to CaCO_3↓ + H_2O$
0,25mol
$n_{CO_2} = \dfrac{5,6}{22,4} = 0,25 (mol)$
$⇒n_{Ca(OH)_2} = \dfrac{0,25 . 1}{1} = 0,25 mol$
$⇒C_{M_{Ca(OH)_2}} = \dfrac{0,25}{0,1} = 2,5 (M)$
$b) n_{CaCO_3} = \dfrac{0,25 .1}{1} = 0,25 mol$
$⇒m_{CaCO_3} = 0,25 . (40 + 12 + 16 .3)=25 (g)$
c) PTHH
$ Ca(OH)_2 + 2HCl \to CaCl_2 + H_2O$
0,25mol
$⇒n_{HCl} = \dfrac{0,25 . 2}{1} = 0,5 (mol)$
$⇒m_{HCl} = 0,5 . (1 + 35,5)= 18,25 (g)$
Ta có :$\text{C%$_{HCl}$=$\dfrac{m_{HCl}}{m_{dd HCl}} . 100$%}$
$⇒m_{dd HCl} = \dfrac{m_{HCl}}{C_{HCl}} . 100$% = $\dfrac{18,25}{20%} . 100$% = $91,25 (g)$