$2HgO\buildrel{{t^o}}\over\to 2Hg+O_2$
a,
$n_{HgO}=\dfrac{21,6}{217}=0,1(mol)$
$\Rightarrow n_{O_2}=\dfrac{1}{2}n_{HgO}=0,05(mol)$
$\Rightarrow V_{O_2}=0,05.22,4=1,12l$
b,
$n_{HgO}=\dfrac{43,2}{217}=0,2(mol)$
$\Rightarrow n_{Hg}=n_{HgO}=0,2(mol)$
$\Rightarrow m_{Hg}=0,2.201=40,2g$
c,
$n_{Hg}=\dfrac{14,07}{201}=0,07(mol)$
$\Rightarrow n_{HgO}=n_{Hg}=0,07(mol)$
$\Rightarrow m_{HgO}=0,07.217=15,19g$
