$H_2+PbO\xrightarrow{{t^o}} Pb+H_2O$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$n_{PbO}=\dfrac{44,6}{223}=0,2(mol)$
$\dfrac{0,1}{1}<\dfrac{0,2}{1}\to PbO$ dư, $H_2$ hết
Theo PTHH:
$n_{Pb}=n_{H_2}=0,1(mol)$
$\to m_{Pb}=0,1.207=20,7g$
$2Pb+O_2\xrightarrow{{t^o}} 2PbO$
Theo PTHH:
$n_{PbO}=b_{Pb}=0,1(mol)$
$\to m_{PbO}=0,1.223=22,3g$