When 0.0500 mol of Au_2S_3 is reduced completely with excess H_2, what mass of Au is produced?

• A. 19.7 g
• B. 9.85 g
• C. 7.87 g
• D. 3.94 g

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation for the reduction of gold(III) sulfide (Au_2S_3) with hydrogen gas (H_2). The reaction produces gold (Au) and hydrogen sulfide (H_2S). The balanced equation is: \(\mathrm{Au_2S_3 + 3H_2 \rightarrow 2Au + 3H_2S}\).
2. Identify the mole ratio between Au_2S_3 and Au from the balanced equation. For every 1 mole of Au_2S_3, 2 moles of Au are produced.
3. Calculate the moles of Au produced by multiplying the moles of Au_2S_3 given (0.0500 mol) by the mole ratio of Au to Au_2S_3: \(\text{moles of Au} = 0.0500 \times 2\).
4. Find the molar mass of gold (Au) using the periodic table. The atomic mass of Au is approximately 197 g/mol.
5. Calculate the mass of Au produced by multiplying the moles of Au by the molar mass of Au: \(\text{mass of Au} = \text{moles of Au} \times 197 \text{ g/mol}\).