Answer :
Answer:
337.272grams
Explanations:
The balanced chemical equation between lead(IV)oxide and hydrochloric acid is given as;
[tex]PbO_2\mleft(s\mright)+4HCl\mleft(aq\mright)\to PbCl_2\mleft(aq\mright)+Cl_2\mleft(g\mright)+2H_2O\mleft(l\mright)[/tex]
From the reaction, we can see that 1 mole of PbO2 produces 1 mole of chlorine gas (Cl2).
Get the number of moles of Chlorine gas;
[tex]\begin{gathered} \text{Moles of Cl}_2=\frac{\text{Mass of chlorine gas}}{Molar\text{ mass of chlorine gas}} \\ \text{Moles of Cl}_2=\frac{100}{2(35.45)} \\ \text{Moles of Cl}_2=\frac{100}{70.9} \\ \text{Moles of Cl}_2=1.410\text{g/mol} \end{gathered}[/tex]Since the stoichiometric ratio of PbO2 and the chlorine gas is 1:1, hence the number of moles of lead(IV) oxide will be 1.410g/mol
Get the molar mass of lead(IV)oxide;
[tex]\text{ Molar mass of PbO}_2=207.2+(2\times16)=239.2\text{g/mol}[/tex]Next is to get the required mass of lead (IV) oxide that is needed to produce 100 grams of chlorine gas as shown;
[tex]\begin{gathered} \text{Mass of PbO}_2=\text{Number of moles }\times\text{ molar mass} \\ \text{Mass of PbO}_2=1.410mol\times\frac{239.2g}{\text{mol}} \\ \text{Mass of PbO}_2=337.272\text{grams} \end{gathered}[/tex]Therefore the mass of lead (IV) oxide needed to produce 100 grams of chlorine gas is 337.272grams
