Đáp án:
1. \(V_{{O_2}}=22,4\) (lít)
2. \(V_{{H_2}}=4,48\) (lít)
Giải thích các bước giải:
\[\begin{array}{l}
1,\\
{d_{A/kk}} = 0,552 \Leftrightarrow {M_A} = 0,552 \cdot 29 = 16{\rm{g/mol}}\\
A:\;{C_x}{H_y}\\
\Rightarrow {M_A} = 12x + y = 16\;(1)\\
\frac{{\% {m_C}}}{{\% {m_H}}} = \frac{{75\% }}{{25\% }}\\
\Rightarrow \frac{{12x}}{y} = 3\\
\Leftrightarrow 12x - 3y = 0\;(2)\\
(1),(2) \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{x = 1}\\
{y = 4}
\end{array}} \right.\\
\Rightarrow A:\;C{H_4}\\
{n_{C{H_4}}} = \frac{{11,2}}{{22,4}} = 0,5\;{\rm{mol}}\\
{\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} \quad C{H_4} + 2{O_2}\xrightarrow{t^\circ}C{O_2} + 2{H_2}O\\
{\rm{mol}}{\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} 0,5 \to {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} 1\\
\Rightarrow {V_{{O_2}}} = 1 \cdot 22,4 = 22,4\left( {lit} \right)\\
2,\\
a.\;Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b.\;{n_{Zn}} = \frac{{13}}{{65}} = 0,2\;{\rm{mol}}\\
\Rightarrow {n_{{H_2}}} = {n_{Zn}} = 0,2\;{\rm{mol}}\\
\Rightarrow {V_{{H_2}}} = 0,2 \cdot 22,4 = 4,48\;\left( {lit} \right)
\end{array}\]
