Question

How much sulphur dioxide is produced on complete combustion of 1 kg of coal containing 6.23percent sulphur?

Answer 1

Answer:

Approximately $1.24 \times 10^{2}\; \rm g$, assuming that all sulfur in that coal was converted to $\rm SO_2$.

Explanation:

Look up the relative atomic mass of $\rm S$ and $\rm O$ on a modern periodic table:

• $\rm S$: $32.06$.
• $\rm O$: $15.999$.

Convert the unit of the mass of coal to grams:

$\begin{aligned} & m(\text{coal})= 1\; \rm kg \times \frac{10^{3}\; \rm g}{1\; \rm kg} = 1000\; \rm g\end{aligned}$.

Mass of sulfur in that much coal:

$m(\text{sulfur}) = 1000\; \rm g \times 6.23\% = 62.3\; \rm g$.

The relative atomic mass of sulfur is $32.06$. Therefore, the mass of each mole of sulfur atoms would be $32.06\; \rm g$. Calculate the number of moles of atoms in that $62.3\; \rm g$ of sulfur:

$\begin{aligned}& n(\text{S})\\&= \frac{m(\mathrm{S})}{M(\mathrm{S})}\\ &= \frac{62.3\; \rm g}{32.06\; \rm g \cdot mol^{-1}} \approx 1.94323\; \rm mol\end{aligned}$.

Each $\rm SO_2\!$ molecule contains one sulfur atom. Therefore, assuming that all those (approximately) $1.94323\; \rm mol\!$ of sulfur atoms were converted to $\rm SO_2$ molecules through the reaction with $\rm O_2$, (approximately) $1.94323\; \rm mol$ of $\!\rm SO_2$ molecules would be produced.

Calculate the mass of one mole of $\rm SO_2$ molecules:

$\begin{aligned}& M(\mathrm{SO_2})\\ &= 32.06 + 2\times 15.999 \\ &= 64.058\; \rm g \cdot mol^{-1}\end{aligned}$.

The mass of that $1.94323\; \rm mol$ of $\rm SO_2$ molecules would be:

$\begin{aligned}& m(\mathrm{SO_2}) \\ &= n(\mathrm{SO_2}) \cdot M(\mathrm{SO_2}) \\ &\approx 19.4323\; \rm mol \\ &\quad \times 64.058\; \rm g \cdot mol^{-1} \\ &\approx 1.24\times 10^{2}\; \rm g\end{aligned}$.