Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 97

See the answer below.

Number of moles of $KClO_4$: $234\ kg\times 1000\ g/kg\div 138.55\ g/kmol= 1689\ mol$ From stoichiometry: $1689\ mol\times 4\ mol\ KClO_3/3\ mol\ KClO_4\times 3\ mol\ KClO/1\ mol\ KClO_3\times 1\ mol\ Cl_2/1\ mol\ KClO\times 70.90\ g/mol\ Cl_2\div 1000\ g/kg=479.0\ kg\ Cl_2$