Question

How many moles of Lead ($\mathrm{II}$) chloride will be formed from a reaction between $6.5\mathrm{g}\mathrm{PbO}$ and $3.2\mathrm{g}\mathrm{HCl}$?

Answer 1

Step 1: Given data

1. Given the mass of $\mathrm{PbO}=6.5\mathrm{gm}$
2. Given the mass of $\mathrm{HCl}=3.2\mathrm{gm}$

Step 2: Calculating moles of lead(II) Chloride:

When Lead oxide reacts with Hydrochloric acid there is the formation of Lead ($\mathrm{II}$) chloride $\underset{\mathrm{Lead}\mathrm{oxide}}{\mathrm{PbO}\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\left(\mathrm{II}\right)\mathrm{chloride}}{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)}$

$\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{PbO}=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{Pb}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O}$

$=207+16\Rightarrow 223\mathrm{gm}$

$\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{HCl}=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{H}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{Cl}$

$=1+35.5\Rightarrow 36.5\mathrm{gm}$

The number of moles is calculated as

Number of moles $=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}$

Number of moles of$\mathrm{PbO}=\frac{6.5}{223}=0.029$

Number of moles of $\mathrm{HCl}=\frac{3.2}{36.5}=0.0877$

From the number of moles calculations, $\mathrm{PbO}$ is the limiting agent as its moles are less than $\mathrm{HCl}$.

When one mole of $\mathrm{PbO}$ is totally consumed there is the formation of one mole of ${\mathrm{PbCl}}_2$.

Hence, 0.029 moles of Lead(II) chloride are formed.