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

[tommyjohn]

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

 

[Borek]

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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[Blueshift5]

where to start?

I would approach this problem by first identifying the limiting reactant. This is the reactant that will be completely used up in the reaction and will determine the amount of product that can be formed. In this case, we have three reactants: Ag, H(2)S, and O(2). To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric coefficients in the balanced equation.

19.00g Ag is equivalent to 0.127 moles of Ag (using the molar mass of Ag, which is 107.87 g/mol).
@.80 H(2)S is equivalent to 0.010 moles of H(2)S (assuming that the @ symbol is a typo and should be a number).
1.600g O(2) is equivalent to 0.050 moles of O(2) (using the molar mass of O(2), which is 32.00 g/mol).

Next, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation. From the equation, we can see that 4 moles of Ag react with 2 moles of H(2)S and 1 mole of O(2). This means that 0.127 moles of Ag would react with 0.0635 moles of H(2)S and 0.03175 moles of O(2). Since we have more than 0.0635 moles of H(2)S and 0.03175 moles of O(2), these reactants are not limiting.

However, we only have 0.010 moles of H(2)S, which is less than the amount needed for the reaction. This means that H(2)S is the limiting reactant. Therefore, we can only form 0.005 moles of Ag(2)S from the reaction.

To calculate the mass of Ag(2)S formed, we can use the molar mass of Ag(2)S, which is 247.8 g/mol. Multiplying 0.005 moles by the molar mass, we get 1.239 g of Ag(2)S.

To determine the amount of non-limiting reactants left, we can use the same method as above