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

Question

Aicharya · Accepted Answer

Key Idea : The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles $=\frac{\text { weight }}{\text { molecular weight }}$
$\begin{array}{ccc}
\mathrm{PbO} & 2 \mathrm{HCl} \longrightarrow & \mathrm{PbCl}_2+\mathrm{H}_2 \mathrm{O} \\
207.2+16 & 2(35.5+1) & 207.2+71 \\
=223.2 & =73 & =278.2
\end{array}$
Here, 1 mole of $\mathrm{PbO}$ reacts with 2 moles of $\mathrm{HCl}$, thus $\mathrm{PbO}$ is the limiting reagent
$\because 223.2 \mathrm{~g}$ PbO gives $\mathrm{PbCl}_2=278.2 \mathrm{~g}$
$\therefore 6.5 \mathrm{~g} \mathrm{PbO}$ will give $\mathrm{PbCl}_2$
$\begin{aligned}
& =\frac{278.2}{223.2} \times 6.5 \mathrm{~g} \\
& =\frac{278.2 \times 6.5}{223.2 \times 278.2} \mathrm{~mol} \\
& =0.029 \mathrm{~mol}
\end{aligned}$

How many moles of lead (II) chloride will be formed from a reaction between $6.5 \mathrm{~g}$ of…

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