Question

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

• A. $0.011$
• B. $0.029$
• C. $0.044$
• D. $0.333$

Answer 1

The correct option is B $0.029$

Moles of $PbO$$=\frac{m_{PbO}}{M{M}_{PbO}}=\frac{6.5}{223}=0.029mol$

Moles of $HCl$$=\frac{3.2}{36.5}=0.0877mol$
The reaction is,
$PbO+2HCl\to PbC{l}_2+H_{2}O$
From the chemical reaction above, $1mol$ of $PbO$ reacts completely with $2mol$ of $HCl$.
Therefore, $0.029mol$ of $PbO$ reacts completely with $0.058mol$ of $HCl$.
Thus, $PbO$ will act as limiting reagent and will decide the amount of $PbC{l}_2$ formed.
So, moles of $PbC{l}_2$ formed $=\frac{1\times 0.029}{1}=0.029mol$

Hence the correct answer is option (b).