$2HgO\buildrel{{t^o}}\over\to 2Hg+O_2$

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,

$n_{Hg}=\dfrac{14,07}{201}=0,07(mol)$

$\Rightarrow n_{HgO}=n_{Hg}=0,07(mol)$

$\Rightarrow m_{HgO}=0,07.217=15,19g$

Đáp án:

$2HgO\rightarrow 2Hg+O_2$

a)

$n_{HgO}=\dfrac{21,6}{217}=0,1(mol)$

$n_{O_2}=\dfrac{1}{2}n_{HgO}=0,05(mol)$

$V_{O_2}=0,05.22,4=1,12(l)$

b)

$n_{HgO}=\dfrac{43,2}{217}=0,2(mol)$

$n_{Hg}=n_{HgO}=0,2(mol)$

$m_{Hg}=0,2.201=40,2(g)$

c)

$n_{Hg}=\dfrac{14,07}{201}=0,07(mol)$

$n_{HgO}=n_{Hg}=0,07(mol)$

$m_{HgO}=0,07.217=15,19(g)$