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2 years ago

11

Sodium sulfate reacts with carbon to produce the products sodium sulfide and carbon monoxide. Identify the reducing agent in the

following reaction.
Na2SO4 + 4C yields Na2S + 4CO

Na2SO4
C
Na2S
CO

2 answers:

Lunna [17]2 years ago

7 0

Answer : The correct option is, carbon (C)

Explanation :

Oxidizing agent : Oxidizing agent are the agent that gains electrons and is reduced in a chemical reaction.

Reducing agent : Reducing agent are the agent that lose electrons and is oxidized in a chemical reaction.

The given balanced chemical reaction is,

$Na_2SO_4+4C\rightarrow Na_2S+4CO$

First we have to calculate the oxidation number of sulfur and carbon.

The oxidation state of 'S' in $Na_2SO_4$ and $Na_2S$ are, (+6) and (-2) respectively.

The oxidation state of 'C' in $4C$ and $4CO$ are, (0) and (+2) respectively.

From this we conclude that, the $Na_2SO_4$ act as an oxidizing agent and 'C' act as a reducing agent.

Hence, the reducing agent in the following reaction is, carbon (C)

Maslowich2 years ago

6 0

The reducing agent will itself be oxidized.
The oxidation number of carbon goes form 0 to +2. Therefore, it is the reducing agent.

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Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

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In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
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Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
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Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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