Chemistry, 21.12.2020 14:00, edtorres5

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

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answered: jonesmegan526

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:

$\rm S$ :

. $\rm O$ : 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:

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

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

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