Question

Answer 1

a) PbO + H₂ --t^o--> Pb + H₂O

b) \(n_{PbO}=\dfrac{60}{223}\left(mol\right)\)

Theo PTHH: \(n_{Pb}=\dfrac{60}{223}\left(mol\right)\Rightarrow m_{Pb}=\dfrac{60}{223}.207=\dfrac{12420}{223}\left(g\right)\)

c) Theo PTHH: \(n_{H_2}=\dfrac{60}{223}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{60}{223}.22,4=\dfrac{1344}{223}\left(l\right)\)

Answer 2

\(n_{PbO}=\dfrac{60}{223}=0,26\left(mol\right)\)

\(PbO+H_2\rightarrow\left(t^o\right)Pb+H_2O\)

0,26     0,26         0,26              ( mol )

\(m_{Pb}=0,26.207=53,82\left(g\right)\)

\(V_{H_2}=0,26.22,4=5,824\left(l\right)\)