## Chemistry and Chemical Reactivity (9th Edition)

$0.1479\ kg\ of \ Sb_2S_3$
Atomic weights (g/mol): $Sb: 121.760,\ S: 32.06$ Molar mass: $339.70\ g/mol$ Mass fraction of Sb: $2\times121.760/339.70\times100\%=71.69\%$ For 1.00 kg of the ore with 10.6% Sb: $1.00\times10.6/100=1.00\times[71.69/100\times x +(1-x)\times0]$ $x=14.79\%$ In 1.00 kg of the ore, there's $0.1479\ kg\ of \ Sb_2S_3$