High School

What mass of Au is produced when 0.0500 mol of [tex]$Au_2S_3$[/tex] is reduced completely with excess [tex]$H_2$[/tex]?

A. 9.85 g
B. 19.7 g
C. 24.5 g
D. 39.4 g
E. 48.9 g

Answer :

To solve this problem, we need to determine the mass of gold (Au) produced when 0.0500 moles of gold(III) sulfide ([tex]\(Au_2S_3\)[/tex]) is completely reduced with excess hydrogen ([tex]\(H_2\)[/tex]).

### Step-by-Step Solution:

1. Understand the Reaction:
The chemical equation for the reduction of [tex]\(Au_2S_3\)[/tex] with hydrogen can be represented as:
[tex]\[
Au_2S_3 + 3H_2 \rightarrow 2Au + 3H_2S
\][/tex]
From this balanced equation, we can see that 1 mole of [tex]\(Au_2S_3\)[/tex] yields 2 moles of Au.

2. Calculate the Moles of Au:
We start with 0.0500 moles of [tex]\(Au_2S_3\)[/tex]. Since 1 mole of [tex]\(Au_2S_3\)[/tex] produces 2 moles of Au, the moles of Au produced will be:
[tex]\[
0.0500 \, \text{mol} \times 2 = 0.100 \, \text{mol Au}
\][/tex]

3. Determine the Molar Mass of Au:
The molar mass of Au (gold) is approximately 197.0 grams per mole (g/mol).

4. Calculate the Mass of Au Produced:
To find the mass of Au produced, we use the formula:
[tex]\[
\text{Mass} = \text{moles} \times \text{molar mass}
\][/tex]
Substituting the known values:
[tex]\[
\text{Mass of Au} = 0.100 \, \text{mol} \times 197.0 \, \text{g/mol} = 19.7 \, \text{g}
\][/tex]

Thus, the mass of gold produced is 19.7 grams. The correct answer is option B.

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