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What mass of Au is produced when 0.0500 mol of [tex]\text{Au}_2\text{S}_3[/tex] is reduced completely with excess [tex]\text{H}_2[/tex]?

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

Answer :

To solve the problem of determining the mass of gold (Au) produced when 0.0500 moles of Au₂S₃ are reduced completely with excess hydrogen (H₂), we can follow these steps:

1. Understand the Chemical Reaction: The compound Au₂S₃ decomposes to produce gold (Au) and sulfur (S) when reduced with hydrogen. The balanced chemical equation for the reaction can be written as:

[tex]\( \text{Au}_2\text{S}_3 + \text{H}_2 \rightarrow 2\text{Au} + \text{Sulfur Compounds} \)[/tex]

From the stoichiometry of the equation, 1 mole of Au₂S₃ produces 2 moles of Au.

2. Calculate Moles of Au Produced: If you start with 0.0500 moles of Au₂S₃, then the moles of Au produced can be calculated using the stoichiometry:

[tex]\[ \text{Moles of Au} = 2 \times \text{Moles of Au₂S₃} = 2 \times 0.0500 \, \text{moles} = 0.100 \text{ moles of Au} \][/tex]

3. Molar Mass of Au: The molar mass of gold (Au) is approximately 196.97 grams per mole.

4. Calculate the Mass of Au Produced: To find the mass of gold produced, use the formula:

[tex]\[ \text{Mass of Au} = \text{Moles of Au} \times \text{Molar Mass of Au} \][/tex]
[tex]\[ \text{Mass of Au} = 0.100 \, \text{moles} \times 196.97 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of Au} = 19.697 \, \text{g} \][/tex]

Therefore, the mass of gold produced when 0.0500 moles of Au₂S₃ are reduced completely is approximately 19.7 grams. The closest option that matches this result is 19.7 g.

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