Mole Concept and Stoichiometry — Question 3

(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