PTHH : \(HgO+H_2\underrightarrow{t^o}Hg+H_2O\)
a) \(n_{HgO}=\dfrac{m}{M}=\dfrac{21,7}{217}=0,1\left(mol\right)\)
Ta có : 1mol-----------1mol
Suy ra : 0,1mol--------0,1mol
\(\Rightarrow m_{Hg}=n.M=0,1.201=20,1\left(g\right)\)
b) Ta có : \(n_{H_2}=0,1mol\)
\(\Rightarrow V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)