Answer:
Approximately [tex]1.24 \times 10^{2}\; \rm g[/tex], assuming that all sulfur in that coal was converted to [tex]\rm SO_2[/tex].
Explanation:
Look up the relative atomic mass of [tex]\rm S[/tex] and [tex]\rm O[/tex] on a modern periodic table:
Convert the unit of the mass of coal to grams:
[tex]\begin{aligned} & m(\text{coal})= 1\; \rm kg \times \frac{10^{3}\; \rm g}{1\; \rm kg} = 1000\; \rm g\end{aligned}[/tex].
Mass of sulfur in that much coal:
[tex]m(\text{sulfur}) = 1000\; \rm g \times 6.23\% = 62.3\; \rm g[/tex].
The relative atomic mass of sulfur is [tex]32.06[/tex]. Therefore, the mass of each mole of sulfur atoms would be [tex]32.06\; \rm g[/tex]. Calculate the number of moles of atoms in that [tex]62.3\; \rm g[/tex] of sulfur:
[tex]\begin{aligned}& n(\text{S})\\&= \frac{m(\mathrm{S})}{M(\mathrm{S})}\\ &= \frac{62.3\; \rm g}{32.06\; \rm g \cdot mol^{-1}} \approx 1.94323\; \rm mol\end{aligned}[/tex].
Each [tex]\rm SO_2\![/tex] molecule contains one sulfur atom. Therefore, assuming that all those (approximately) [tex]1.94323\; \rm mol\![/tex] of sulfur atoms were converted to [tex]\rm SO_2[/tex] molecules through the reaction with [tex]\rm O_2[/tex], (approximately) [tex]1.94323\; \rm mol[/tex] of [tex]\!\rm SO_2[/tex] molecules would be produced.
Calculate the mass of one mole of [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}& M(\mathrm{SO_2})\\ &= 32.06 + 2\times 15.999 \\ &= 64.058\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The mass of that [tex]1.94323\; \rm mol[/tex] of [tex]\rm SO_2[/tex] molecules would be:
[tex]\begin{aligned}& m(\mathrm{SO_2}) \\ &= n(\mathrm{SO_2}) \cdot M(\mathrm{SO_2}) \\ &\approx 19.4323\; \rm mol \\ &\quad \times 64.058\; \rm g \cdot mol^{-1} \\ &\approx 1.24\times 10^{2}\; \rm g\end{aligned}[/tex].
