Đáp án:
b) mCaCO3=5 g; mNaCl = 3,8 g
c) 12,3ml
Giải thích các bước giải:
a) $CaC{O_3} + 2C{H_3}COOH \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O$
b)
$\begin{gathered}
{n_{C{O_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol \Rightarrow {n_{CaC{O_3}}} = 0,05mol \hfill \\
\Rightarrow {m_{CaC{O_3}}} = 0,05.100 = 5(g) \hfill \\
\Rightarrow {m_{NaCl}} = 8,8 - 5 = 3,8\left( g \right) \hfill \\
\end{gathered} $
c) ${n_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 0,1mol$
${C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O$
$\begin{gathered}
H = 80\% \hfill \\
{n_{{C_2}{H_5}OH}} = \dfrac{{{n_{C{H_3}COOH}}}}{{0,8}} = 0,125mol \Rightarrow {m_{{C_2}{H_5}OH}} = 0,125.46 = 5,75\left( g \right) \hfill \\
\Rightarrow {V_{{C_2}{H_5}OH}} = \frac{m}{D} = \dfrac{{5,75}}{{0,78}} = 7,37ml \hfill \\
\Rightarrow {V_{dd{C_2}{H_5}OH}} = \dfrac{{7,37.100}}{{60}} = 12,3ml \hfill \\
\end{gathered} $