Some Basic Concepts of Chemistry - Result Question 47
####47. How many moles of lead (II) chloride will be formed from a reaction between $6.5 g$ of $PbO$ and $3.2 g$ of $HCl$ ?
(a) 0.044
(b) 0.333
(c) 0.011
(d) 0.029
[2008]
Show Answer
Solution:
- (d) Writing the equation for the reaction, we get
$ \begin{aligned} & PbO+2 HCl \longrightarrow PbCl_2+H_2 O \\ & 207+16 \quad 2 \times 36.5 \quad 207+71 \\ & =223 g \quad=73 g \quad=278 g \end{aligned} $
From this equation we find $223 g$ of $PbO$ reacts with $73 g$ of $HCl$ to form $278 g$ of $PbCl_2$.
If we carry out the reaction between $3.2 g HCl$ and $6.5 g PbO$.
Amount of $PbO$ that reacts with $3.2 g HCl$ $=\frac{223}{73} \times 3.2=9.77 g$.
Since amount of $PbO$ present is only $6.5 g$ so $PbO$ is the limiting reagent.
Amount of $PbCl_2$ formed by $6.5 g$ of $PbO$
$=\frac{278}{223} \times 6.5 g$
Number of moles of $PbCl_2$ formed
$=\frac{278}{223} \times \frac{6.5}{278}$ moles $=0.029$ moles.
