a,

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$HgO+H_2\xrightarrow{{t^o}} Hg+H_2O$

$\to n_{Hg}=n_{H_2}=0,3(mol)$

Chất rắn là thuỷ ngân.

$\to m=m_{Hg}=0,3.201=60,3g$

b,

$n_{O_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$2H_2+O_2\xrightarrow{{t^o}} 2H_2O$

$\dfrac{0,1}{1}<\dfrac{0,3}{2}$

$\to O_2$ hết, $H_2$ dư 

Vậy sau phản ứng có $H_2$, $H_2O$

$n_{H_2\text{pứ}}=n_{H_2O}=2n_{O_2}=0,2(mol)$

$\to m_{H_2O}=0,2.18=3,6g$

$n_{H_2\text{dư}}=0,3-0,2=0,1(mol)$

$\to m_{H_2}=0,1.2=0,2g$

$\rm PTHH:\\a) \\ HgO + H_2 \to Hg + H_2O\\n_{H_2} = \dfrac{6,72}{22,4} = 0,3\ mol\\ Theo\ PT\ ta\ có\ n_{Hg} = n_{H_2} = 0,3\ mol\\ \to m\ = m_{Hg} = 0,3 * 201 = 60,3 g\\ b) 2H_2 + O_2 \to 2H_2O \\ n_{O_2} = \dfrac{2,24}{22,4} = 0,1\ mol \\Ta\ có\ tỉ\ lệ\  \dfrac{0,3}{2} > \dfrac{0,1}{1} \\ \to H_2\ dư \\Theo\ PT\ ta\ có\\ n_{H_2}\ (phản ứng)\ = 2n_{O_2} = 2*0,1 = 0,2\ mol \\ \to n_{H_2}\ (dư)\ = 0,3 - 0,2 = 0,1\ mol\\ \to m_{H_2}\ (dư)\ = 0,1 * 2 = 0,2g\\ Theo\ PT\ có\ n_{H_2O} = n_{H_2}\ (phản ứng)\ = 0,2\ mol\\ \to m_{H_2O} = 0,2 *18 = 3,6g$