Question

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

​

Answer 1

Approximately
`1.24 * 10^(2)\; \rm g`, assuming that all sulfur in that coal was converted to
`\rm SO_2`.

Step-by-step 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 * (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 * 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})}\\ &= (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* 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 * 64.058\; \rm g \cdot mol^(-1) \\ &\approx 1.24* 10^(2)\; \rm g\end{aligned}`.