Đáp án:

$\rm 68,41\%$

Giải thích các bước giải:

$\rm PbO+H_2\xrightarrow{t^o}Pb+H_2O$

Bảo toàn KL:

$\rm m_{PbO}+m_{H_2}=m_{CR}+m_{H_2O}$

$\rm \to m_{H_2O}-m_{H_2}=55,75-52,95=2,8(g)$

$\rm \to 18n_{H_2O}-2n_{H_2}=2,8$

Mà $\rm n_{H_2O}=n_{H_2}$

$\rm \to 16n_{H_2}=2,8$

$\rm \to n_{H_2O}=n_{H_2}=0,175(mol)$

Theo PT: $\rm n_{PbO(pứ)}=n_{Pb}=n_{H_2}=0,175(mol)$

$\rm \to m_{PbO(pứ)}=0,175.223=39,025(g)$

$\rm \to m_{PbO(dư)}=55,75-39,025=16,725(g)$

$\rm \to m_{Pb}=0,175.207=36,225(g)$

$\rm \to \%m_{Pb(trong\,X)}=\dfrac{36,225}{36,225+16,725}.100\%\approx 68,41\%$