Đáp án:
\(\begin{array}{l}
a)\\
\% Zn = 64,4\% \\
\% FeO = 35,6\% \\
b)\\
{m_{ddC{H_3}COOH}} = 36g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
FeO + 2C{H_3}COOH \to {(C{H_3}COO)_2}Fe + {H_2}O\\
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,448}}{{22,4}} = 0,02mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,02mol\\
{m_{Zn}} = 0,02 \times 65 = 1,3g\\
\% Zn = \dfrac{{1,3}}{{2,02}} \times 100\% = 64,4\% \\
\% FeO = 100 - 64,4 = 35,6\% \\
b)\\
{m_{FeO}} = 2,02 - 1,3 = 0,72g\\
{n_{FeO}} = \dfrac{{0,72}}{{72}} = 0,01mol\\
{n_{C{H_3}COOH}} = 2{n_{Zn}} + 2{n_{FeO}} = 0,06mol\\
{m_{C{H_3}COOH}} = 0,06 \times 60 = 3,6g\\
{m_{ddC{H_3}COOH}} = \dfrac{{3,6 \times 100}}{{10}} = 36g
\end{array}\)
