Copper is extracted from 1.00 tonne of an ore containing 55.0% by mass of chalcopyrite, CuFeS2, by roasting, smelting and bessemerisation. Assume…

The chalcopyrite present is $0.550\times1000\ \mathrm{kg}=550\ \mathrm{kg}$. Its molar mass is $63.55+55.85+2(32.06)=183.52\ \mathrm{g\,mol^{-1}}$, so $550\times10^{3}/183.52=2996.9\ \mathrm{mol}$. The processed amount is $0.900\times2996.9=2697.3\ \mathrm{mol}$. In the sequence $\mathrm{2CuFeS_2+O_2\to Cu_2S+2FeS+SO_2}$, then $\mathrm{2FeS+3O_2\to2FeO+2SO_2}$ with $\mathrm{FeO+SiO_2\to FeSiO_3}$, and then $\mathrm{2Cu_2S+3O_2\to2Cu_2O+2SO_2}$ followed by $\mathrm{Cu_2S+2Cu_2O\to6Cu+SO_2}$, all the sulfur leaves as $\mathrm{SO_2}$. Each $\mathrm{CuFeS_2}$ has $2$ S, so it gives $2$ $\mathrm{SO_2}$. Moles of $\mathrm{SO_2}=2\times2697.3=5394.6$. With $M(\mathrm{SO_2})=64.06$ the mass is $5394.6\times64.06=345{,}572\ \mathrm{g}=345.57\ \mathrm{kg}$. As a cross-check, the copper recovered would be $2697.3\times63.55=171.4\ \mathrm{kg}$, while the sulfur dioxide mass is almost exactly twice that, because each copper atom is accompanied by two sulfur atoms and sulfur dioxide has a mass of $64.06$ against $63.55$ for copper.