Question

The major product formed when $P{b}_3{O}_4$ reacts with conc. $HCl$ is $PbC{l}_2$.
If true enter 1, else enter 0.

Answer 1

The reaction of $P{b}_3{O}_4$ with conc. $HCl$ is as follows:

$P{b}_3{O}_4+8HCl\to PbC{l}_2+4H_{2}O+C{l}_2$

The major product formed is $PbC{l}_2$.

Hence, the statement is true.