Đáp án:
\(\begin{array}{l}
b)\\
x = 13,04g\\
c)\\
{V_{{O_2}}} = 34,496l\\
{V_{kk}} = 172,48l\\
d)\\
{m_{{H_2}O}} = 24,48g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O(1)\\
2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O(2)\\
b)\\
{n_{C{H_4}}} = \dfrac{{2,24}}{{16}} = 0,14\,mol\\
{n_{C{O_2}(1)}} = {n_{C{H_4}}} = 0,14\,mol\\
{n_{C{O_2}}} = \dfrac{{19,264}}{{22,4}} = 0,86\,mol\\
\Rightarrow {n_{C{O_2}(2)}} = 0,86 - 0,14 = 0,72\,mol\\
{n_{{C_2}{H_6}}} = \dfrac{{0,72 \times 2}}{4} = 0,36\,mol\\
x = {m_{C{H_4}}} + {m_{{C_2}{H_6}}} = 2,24 + 0,36 \times 30 = 13,04g\\
c)\\
{n_{{O_2}}} = 0,14 \times 2 + \dfrac{{0,36 \times 7}}{2} = 1,54\,mol\\
{V_{{O_2}}} = 1,54 \times 22,4 = 34,496l\\
{V_{kk}} = 34,496 \times 5 = 172,48l\\
d)\\
{n_{{H_2}O}} = 0,14 \times 2 + 0,36 \times 3 = 1,36\,mol\\
{m_{{H_2}O}} = 1,36 \times 18 = 24,48g
\end{array}\)
