# Reaction As2S3 + H+ + NO3- -> NO + H2O + AsO4^(3-) + SO4^(2-) The relationship between the molecular weight (M) and the equivalent weight (E) of As2S3 in this reaction is: A) E = M/2 B) E = M/4 C) E = M/24 D) E = M/28

## Question

### Reaction

$$ \begin{array}{l} \text{As}_2\text{S}_3 + \text{H}^+ + \text{NO}_3^- \rightarrow \text{NO} + \text{H}_2\text{O} + \text{AsO}_4^{3-} + \text{SO}_4^{2-} \end{array} $$

The relationship between the molecular weight (M) and the equivalent weight (E) of As$_2$S$_3$ in this reaction is:

- A $ E = \frac{M}{2} $
- B $ E = \frac{M}{4} $
- C $ E = \frac{M}{24} $
- D $ E = \frac{M}{28} $

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## Answer

To determine the relationship between the **molecular mass** ($M$) and the **equivalent mass** ($E$) of As$_2$S$_3$, consider the given reaction:

$$
\begin{array}{l}
\mathrm{As}*{2} \mathrm{S}*{3}+\mathrm{H}^{+}+\mathrm{NO}*{3}^{-} \rightarrow \mathrm{NO}+\mathrm{H}*{2} \mathrm{O}+\mathrm{AsO}*{4}^{3-} +
\mathrm{SO}*{4}^{2-}
\end{array}
$$

For As$_2$S$_3$, let's analyze the equivalence factor:

- The factor $n$ is calculated using the number of atoms and their oxidation states.
- For each atom in As$_2$S$_3$, the oxidation state changes need to be taken into account.

The calculation is as follows:

$$ n\text{-factor} = |2 \times 2 + 8 \times 3| = 28 $$

So, the equivalent mass ($E$) is related to the molecular mass ($M$) by the equation:

$$ E = \frac{M}{28} $$

Therefore, the correct answer is:

**D. $E = \frac{M}{28}$**

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