Đáp án:
\(\begin{array}{l}
a)\\
{m_{{O_2}}} = 64g\\
{V_{{O_2}}} = 44,8l\\
b)\\
{m_{C{O_2}}} = 44g\\
{m_{{H_2}O}} = 36g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{m}{M} = \dfrac{{16}}{{16}} = 1mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} = 2mol\\
{m_{{O_2}}} = n \times M = 2 \times 32 = 64g\\
{V_{{O_2}}} = n \times 22,4 = 2 \times 22,4 = 44,8l\\
b)\\
{n_{C{O_2}}} = {n_{C{H_4}}} = 1mol\\
{m_{C{O_2}}} = n \times M = 1 \times 44 = 44g\\
{n_{{H_2}O}} = 2{n_{C{H_4}}} = 2mol\\
{m_{{H_2}O}} = n \times M = 2 \times 18 = 36g
\end{array}\)