Đáp án:
a) 0,56l
b) 3,1g
c) 2,04g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4Na + {O_2} \xrightarrow{t^0} 2N{a_2}O\\
{n_{Na}} = \dfrac{m}{M} = \dfrac{{2,3}}{{23}} = 0,1\,mol\\
Theo\,pt:{n_{{O_2}}} = \dfrac{{{n_{Na}}}}{4} = \dfrac{{0,1}}{4} = 0,025\,mol\\
{V_{{O_2}}} = 0,025 \times 22,4 = 0,56l\\
b)\\
Theo\,pt:{n_{N{a_2}O}} = \dfrac{{{n_{Na}}}}{2} = 0,05\,mol\\
{m_{N{a_2}O}} = 0,05 \times 62 = 3,1g\\
c)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{2}{3} \times {n_{{O_2}}} = 0,025 \times \dfrac{2}{3} = \dfrac{1}{{60}}\,mol\\
{m_{KCl{O_3}}} = \dfrac{1}{{60}} \times 122,5 = 2,04g
\end{array}\)
