Đáp án+Giải thích các bước giải:

 `CaCO_3 + 2HCl -> CaCl_2 + CO_2 + H_2O`

a) `n_{CO_2} = (6,1975)/(24,79) = 0,25` (mol)

`n_{CaCO_3} = n_{CO_2} = 0,25` (mol)

`n_{HCl} = 2n_{CO_2} = 2.0,25 = 0,5` (mol)

`->` `m_{CaCO_3} = 0,25.100 = 25` (g)

`m_{HCl} = 0,5.36,5 = 18,25` (g)

`->` `a = m_{dd\ HCl} = 18,25 : 7,3% = 250` (g)

b) `n_{CaCl_2} = n_{CO_2} = 0,25` (mol)

`m_{dd\ sau\ pư} = 25 + 250 - 0,25.44 = 264` (g)

`->` `C%_{CaCl_2} = (0,25.111)/264 .100 ≈ 10,51%`

Lời giải:

$\rm a)\\n_{CO_2}=\dfrac{6,1975}{24,79}=0,25\ (mol)$

Phản ứng xảy ra:

$\rm CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\\kern,2cm0,25\kern,2cm\leftarrow\kern,2cm0,5\kern,1cm\leftarrow\kern,2cm0,25\leftarrow\kern,0cm0,25\kern1cm(mol)$

$\rm m=0,25\cdot 100=25\ (g)\\a=\dfrac{0,5\cdot 36,5}{7,3\%}=250\ (g)$

$\rm b)\\C\%_X=C\%_{CaCl_2}=\dfrac{0,25\cdot 111}{25+250-0,25\cdot 44}\cdot 100\%=10,51\%$