Question

The reaction between red lead and hydrochloric acid is given below:
$P{b}_3{O}_4+8HCLI\to 3PbC{l}_2+4H_{2}O+C{l}_2$
calculate : (a) the amss fo lead chloride formed by the action of 6.85 g of red lead, (b) the mass fo chlorine and (c) the volume of chlorine evolved at STP.

Answer 1

(a) .
Molar mass of $P{b}_3{O}_4$ = 685 g
Molar Mass of $PbC{l}_2$ = 278 g
3 $\times$ 278 g of $PbC{l}_2$ is formed when 685 g of red lead is reacted.
the amount of lead chloride formed by the action of 6.85 g of red lead is $\frac{3\times 278}{685}\times 6.85$ = 8.34 g

(b) moles of $P{b}_3{O}_4$ = $\frac{6.85}{685}$ = 0.01
1 mole of $P{b}_3{O}_4$ gives 1 mole of chlorine.
0.01 mole of $P{b}_3{O}_4$ gives 0.01 mole of chlorine
mass of chlorine = 0.01 $\times$ 71 = 0.71 g | Molar mass of chlorine = 71 g

(c) 1 mol of chlorine = 22.4 l
0.01 mol of chlorine = 0.01 $\times$ 22.4 = 0.224 l