Đáp án:

$\rm a)\\n_{CH_3COOH}=\dfrac{24}{60}=0,4\ (mol)$

$\rm C_{M\ CH_3COOH}=\dfrac{0,4}{50×10^{-3}}=8\ (M)$

$\rm m_{dd\ CH_3COOH}=50×1,2=60\ (g)$

$\rm C\%_{CH_3COOH}=\dfrac{24}{60}×100\%=40\%$

$\rm b)\\n_{CaCO_3}=\dfrac{10×80\%}{100}=0,08\ (mol)\\n_{Fe_2O_3}=\dfrac{10×20\%}{160}=0,0125\ (mol)$

Phản ứng xảy ra:

$\rm 2CH_3COOH+CaCO_3\to (CH_3COO)_2Ca+CO_2+H_2O\\6CH_3COOH+ Fe_2O_3\to 2(CH_3COO)_3Fe+ 3H_2O$

Theo phương trình:

$\rm n_{CH_3COOH}=2n_{CaCO_3}+6n_{Fe_2O_3}=0,235\ (mol)$

$\rm m_{dd\ CH_3COOH}=\dfrac{0,235×60}{40\%}=35,25\ (g)$

`n_{CH_3COOH}=24/60=0,4(mol)`

`a.m_{dd CH_3COOH}=D.V=1,2.50=60(g)`

`C% CH_3COOH=24/60 .100%=40%`

`C_M CH_3COOH={0,4}/{0,5}=0,8(M)`

`b.m_{CaCO_3}=10.80%=8(g)->n_{CaCO_3}=8/100=0,08(mol)`

`m_{Fe_2O_3}=10-8=2(g)->n_{Fe_2O_3}=2/160=0,0125(mol)`

`PTHH:CaCO_3+2CH_3COOH->(CH_3COO)_2Ca+CO_2+H_2O`

`(mol)`  `0,08->`        `0,16`

           `Fe_2O_3+6CH_3COOH->2(CH_3COO)_3Fe+3H_2O`

`(mol)`   `0,0125->`   `0,075`

`m_{CH_3COOH}=(0,16+0,075).60=14,1(g)`

`m_{dd CH_3COOH}={14,1.100}/40 =35,25(g)`