Đáp án:
\(\begin{array}{l}
b)\\
{m_{MgC{O_3}}} = 16,8g\\
{m_{MgO}} = 4g\\
c)\\
C{\% _m} = 11,45\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
MgO + 2C{H_3}COOH \to {(C{H_3}COO)_2}Mg + {H_2}O\\
MgC{O_3} + 2C{H_3}COOH \to {(C{H_3}COO)_2}Mg + C{O_2} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{MgC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
{m_{MgC{O_3}}} = 0,2 \times 84 = 16,8g\\
{m_{MgO}} = 20,8 - 16,8 = 4g\\
c)\\
{n_{MgO}} = \dfrac{4}{{40}} = 0,1mol\\
{n_{C{H_3}COOH}} = 2{n_{MgC{O_3}}} + 2{n_{MgO}} = 0,6mol\\
{m_{C{H_3}COOH}} = 0,6 \times 60 = 36g\\
{m_{ddC{H_3}COOH}} = \dfrac{{36 \times 100}}{{10}} = 360g\\
{m_{ddspu}} = 20,8 + 360 - 0,2 \times 44 = 372g\\
{n_{{{(C{H_3}COO)}_2}Mg}} = {n_{MgO}} + {n_{MgC{O_3}}} = 0,3mol\\
{m_{{{(C{H_3}COO)}_2}Mg}} = 0,3 \times 142 = 42,6g\\
C{\% _m} = \dfrac{{42,6}}{{372}} \times 100\% = 11,45\%
\end{array}\)