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 :

Final answer:

To find the mass of Au produced, balance the equation and use stoichiometry. 0.0500 mol of Au₂S₃ would produce 19.7 g of Au.

Explanation:

In order to determine the mass of Au produced, we need to balance the equation for the reaction and use stoichiometry. The balanced equation for the reaction is:

Au₂S₃ + 6H₂ → 2Au + 3H₂S

From the balanced equation, we can see that 1 mol of Au₂S₃ produces 2 mol of Au. Therefore, for 0.0500 mol of Au₂S₃, the mass of Au produced would be:

(0.0500 mol Au₂S₃) x (2 mol Au / 1 mol Au₂S₃) x (197 g Au / 1 mol Au) = 19.7 g Au

So the correct answer is B. 19.7 g.

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