Question

From this balanced chemical equation, how many grams of lead (IV) oxide are needed to producewith 100 grams of chlorine gas? Assume that you have as much HCl as is needed.PbO2(s) + 4HCl(aq) → PbCl2(aq) + Cl2(g) + 2H2O(l)

Answer 1

337.272grams

Explanations:

The balanced chemical equation between lead(IV)oxide and hydrochloric acid is given as;

`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)`

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;

`\begin{gathered} \text{Moles of Cl}_2=\frac{\text{Mass of chlorine gas}}{Molar\text{ mass of chlorine gas}} \\ \text{Moles of Cl}_2=(100)/(2(35.45)) \\ \text{Moles of Cl}_2=(100)/(70.9) \\ \text{Moles of Cl}_2=1.410\text{g/mol} \end{gathered}`

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;

`\text{ Molar mass of PbO}_2=207.2+(2*16)=239.2\text{g/mol}`

Next is to get the required mass of lead (IV) oxide that is needed to produce 100 grams of chlorine gas as shown;

`\begin{gathered} \text{Mass of PbO}_2=\text{Number of moles }*\text{ molar mass} \\ \text{Mass of PbO}_2=1.410mol*\frac{239.2g}{\text{mol}} \\ \text{Mass of PbO}_2=337.272\text{grams} \end{gathered}`

Therefore the mass of lead (IV) oxide needed to produce 100 grams of chlorine gas is 337.272grams