High School

The following reaction is carried out with 45.00 g of [tex]$Na_2S$[/tex] and excess [tex]$AgNO_3$[/tex]. What is the maximum mass produced?

[tex]\[2 \, AgNO_3 + Na_2S \rightarrow Ag_2S + 2 \, NaNO_3\][/tex]

A. [tex]142.9 \, g[/tex]
B. [tex]100.2 \, g[/tex]
C. [tex]14.17 \, g[/tex]
D. [tex]43.13 \, g[/tex]

Answer :

To solve the problem of finding the maximum mass of [tex]\(Ag_2S\)[/tex] produced from the reaction, follow these steps:

1. Write down the balanced chemical equation:

[tex]\[
2 \, \text{AgNO}_3 + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2 \, \text{NaNO}_3
\][/tex]

From the equation, you can see that 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] reacts to produce 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex].

2. Find the molar masses:

- Molar mass of [tex]\( \text{Na}_2\text{S} \)[/tex] is approximately [tex]\( 78.04 \, \text{g/mol} \)[/tex].
- Molar mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] is approximately [tex]\( 247.80 \, \text{g/mol} \)[/tex].

3. Calculate moles of [tex]\( \text{Na}_2\text{S} \)[/tex]:

Using the given mass of [tex]\( \text{Na}_2\text{S} \)[/tex]:

[tex]\[
\text{moles of Na}_2\text{S} = \frac{\text{mass of Na}_2\text{S}}{\text{molar mass of Na}_2\text{S}} = \frac{45.00 \, \text{g}}{78.04 \, \text{g/mol}} \approx 0.5766 \, \text{moles}
\][/tex]

4. Use stoichiometry to find moles of [tex]\( \text{Ag}_2\text{S} \)[/tex]:

According to the reaction, 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] produces 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex]. Therefore:

[tex]\[
\text{moles of Ag}_2\text{S} = 0.5766 \, \text{moles}
\][/tex]

5. Calculate the mass of [tex]\( \text{Ag}_2\text{S} \)[/tex]:

Use the moles of [tex]\( \text{Ag}_2\text{S} \)[/tex] to find its mass:

[tex]\[
\text{mass of Ag}_2\text{S} = \text{moles of Ag}_2\text{S} \times \text{molar mass of Ag}_2\text{S} = 0.5766 \, \text{moles} \times 247.80 \, \text{g/mol} \approx 142.89 \, \text{g}
\][/tex]

Therefore, the maximum mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] produced is approximately 142.9 grams. The correct answer to the question is (A) 142.9 g.

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