Answer
$0.1479\ kg\ of \ Sb_2S_3$
Work Step by Step
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 $0.1479\ kg\ of \ Sb_2S_3$
