Answer :
Final answer:
To find the mass of Au produced, use the stoichiometry of the balanced equation and the molar mass of Au.
Explanation:
To determine the mass of Au produced when 0.0500 mol of Au,S is completely reduced with excess H₂, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction is:
Au,S + H₂ ⟶ Au + H₂S
From the balanced equation, we can see that 1 mol of Au corresponds to 1 mol of Au,S. Therefore, since we have 0.0500 mol of Au,S, we will also have 0.0500 mol of Au produced.
Now, we can calculate the mass of Au produced using its molar mass, which is 197 g/mol:
Mass of Au = (0.0500 mol) x (197 g/mol) = 9.85 g
Therefore, the mass of Au produced when 0.0500 mol of Au,S is completely reduced with excess H₂ is 9.85 g.
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https://brainly.com/question/30218216
The mass of gold produced when 0.0500 mol of Au₂S is reduced with excess hydrogen is 9.85 g after rounding to two decimal places.
The question involves calculating the mass of gold (Au) produced from the reduction of Au₂S (gold sulfide) with hydrogen gas. To determine the mass of Au produced, we use the stoichiometry of the reaction. Given that the molar mass of Au is 196.97 g/mol, we can calculate the mass of gold for 0.0500 mol:
0.0500 mol Au × 196.97 g/mol Au = 9.8485 g Au
Therefore, the mass of gold produced is 9.85 g, rounding off to two decimal places.