Chemistry Problem: Obtaining Ag(2)S from 19.00g Ag, @.80 H(2)S & 1.600g O(2)

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To determine the grams of Ag(2)S produced from 19.00g of Ag, 0.80g of H(2)S, and 1.600g of O(2), the stoichiometry of the reaction must be analyzed. The balanced equation shows that 4 moles of Ag react with 2 moles of H(2)S and 1 mole of O(2) to produce 2 moles of Ag(2)S. Calculating the moles of each reactant reveals the limiting reagent, which will dictate the maximum yield of Ag(2)S. After identifying the limiting reagent, the remaining amounts of the non-limiting reactants can be calculated. This process involves converting grams to moles and applying stoichiometric ratios from the balanced equation.
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Homework Statement



silver tarnishes in the presence of hydrogen sulfide (rotten egg odor), also found in Chinese dry wall and oxygen because of the reaction:

4Ag + 2H(2)S + O(2) --> 2Ag(2)S + 2H(2)O

How many grams of Ag(2)S could be obtained from a mixture of 19.00g Ag, @.80 H(2)S and 1.600g O(2) and how many grams of the non-limiting reactants are left.

Homework Equations


I am completely lost and have no idea how to even start this equation, I need some hekp thanks


The Attempt at a Solution


Not sure
 
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This is a limiting reagent question, just there are three reagents to take into account.

Ignore everything else for now - how many grams of H2S will react with 19.00g of Ag? Do you have that much H2S?

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