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Question
Balance the following equation:
\[\ce{Pb3O4 + HCl->PbCl2 + H2O + Cl2}\]
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Solution
\[\ce{Pb3O4 + HCl->PbCl2 + H2O + Cl2}\]
There are 4 atoms of oxygen in Pb3O4 so make H atoms 8 in HCl.
\[\ce{Pb3O4 + 8HCl->PbCl2 + H2O + Cl2}\]
Multiply PbCl2 by 3 to make Pb atoms equal and H2O by 4.
\[\ce{Pb3O4 + 8HCl->3PbCl2 + 4H2O + Cl2}\]
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