The reaction between red lead and hydrochloric acid is given below:

Pb₃O₄ + 8HCl ⟶ 3PbCl₂ + 4H₂O + Cl₂

Calculate

(a) the mass of lead chloride formed by the action of 6.85 g of red lead,

(b) the mass of the chlorine and

(c) the volume of chlorine evolved at S.T.P.

(a)

$\begin{matrix} \text{Pb}$3\text{O}4 & + &8\text{HCl} & \longrightarrow \ 3(207) + 4(16) & & 8[1 + 35.5] & \ = 621 + 64 & & = 8(36.5) & \ = 685 \text{ g} & & = 292 \text{ g} & \ & & & \ 3\text{PbCl}2 & + & 4\text{H}2\text{O} & + & \text{Cl}_2 \ 3[207 + 2(35.5)] &&&& 2(35.5) \ = 3[207 + 71] &&&& = 71\text{g} \ = 834 \text{ g} \end{matrix}

685 g of Pb₃O₄ gives = 834 g of PbCl₂

∴ 6.85 g of Pb₃O₄ will give

= $\dfrac{834}{685}$ x 6.85 = 8.34 g

(b) 685g of Pb₃O₄ gives = 71g of Cl₂

∴ 6.85 g of Pb₃O₄ will give

= $\dfrac{71}{685}$ x 6.85 = 0.71 g of Cl₂

(c) 685 g of Pb₃O₄ produces 22.4 L of Cl₂

∴ 6.85 g of Pb₃O₄ will produce

$\dfrac{22.4}{685}$ x 6.85 = 0.224 L of Cl₂