a,
$4Na+O_2\buildrel{{t^o}}\over\to 2Na_2O$
$2Ca+O_2\buildrel{{t^o}}\over\to 2CaO$
b,
Gọi a, b là mol Na, Ca.
$\Rightarrow 23a+40b=8,6$ (1)
$n_{Na_2O}= 0,5a$
$n_{CaO}=b$
$\Rightarrow 62.0,5a+56b=11,8$ (2)
(1)(2)$\Rightarrow a=0,2; b=0,1$
$n_{O_2}=\frac{1}{4}n_{Na}+ \frac{1}{2}n_{Ca}= 0,25.0,2+0,5.0,1=0,1 mol$
$\Rightarrow V_{O_2}= 0,1.22,4=2,24l$
c,
$m_{Na}= 0.2.23=4,6g$
$m_{Ca}= 0,1.40=4g$