Đáp án:

$\rm 1.\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\ (mol)$

Phản ứng xảy ra:

$\rm 2CH_3COOH+MgO\to (CH_3COO)_2Mg+H_2O\\2CH_3COOH+MgCO_3\to (CH_3COO)_2Mg+CO_2+H_2O$

Theo phương trình:

$\rm n_{MgCO_3}=n_{CO_2}=0,2\ (mol)$

Khối lượng mỗi chất:

$\rm m_{MgCO_3}=0,2×84=16,8\ (g)\\m_{MgO}=20,8-16,8=4\ (g)$

Phần trăm khối lượng:

$\rm\%m_{MgCO_3}=\dfrac{16,8}{20,8}×100\%=80,8\%\\\%m_{MgO}=(100-80,1)\%=19,2\%$

$\rm 2.\\n_{MgO}=\dfrac{4}{40}=0,1\ (mol)$

Theo phương trình:

$\rm n_{CH_3COOH}=2×(n_{MgCO_3}+n_{MgO})=0,6\ (mol)$

$\rm C\%_{CH_3COOH}=\dfrac{0,6×60}{200}×100\%=18\%$

$\rm 3.$

Bảo toàn khối lượng:

$\rm m_{dd\ sau\ pu}=20,8+200-0,2×44=212\ (g)$

Theo phương trình:

$\rm n_{(CH_3COO)_2Mg}=n_{MgCO_3}+n_{MgO}=0,3\ (mol)$

$\rm C\%_{(CH_3COO)_2Mg}=\dfrac{0,3×142}{212}×100\%=20,1\%$

câu 5:

$\rm n_{CO_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$\rm PTHH:MgO+2CH_3COOH \xrightarrow{} (CH_3COO)_2Mg+H_2O$
$\rm MgCO_3+2CH_3OOH \xrightarrow{} (CH_3COO)_2Mg+CO_2+H_2O$
$\rm 1.n_{MgCO_3}=n_{CO_2}=0,2(mol)$
$\rm m_{MgCO_3}=0,2.84=16,8(g)$

$\rm m_{MgO}= 20,8-14,8=4(g)$

$\rm \%MgO=\dfrac{4.100\%}{20,8}=19,2\%$

$\rm \%MgCO_3=100-19,2=80,8\%$

$\rm 2.n_{MgO}=\dfrac{4}{40}=0,1(mol)$

$\rm n_{CH_3COO}=2n_{MgO}+2n_{MgCO_3}=0,6(mol)$
$\rm C\% CH_3COOH=\dfrac{0,6.60.100\%}{200}=18\%$

$\rm 3.n_{(CH_3COO)_2Mg}=n_{MgO}+n_{MgCO_3}=0,3(mol)$

$\rm m_{dd sau}=20,8+200-0,2.44=212(g)$

$\rm C\% (CH_3COO)_2Mg=\dfrac{0,3.142.100\%}{212}=20,09\%$