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If 5.00 g of hydrosulfuric acid is reacted with 5.00 g of silver nitrate, calculate the mass (in grams) of solid silver sulfide formed.

Answer :

Final answer:

To calculate the mass of solid silver sulfide formed, first determine the limiting reactant between hydrosulfuric acid and silver nitrate. Then use the moles of the limiting reactant to calculate the moles of silver sulfide formed. Finally, multiply the moles of silver sulfide by its molar mass to find the mass.

Explanation:

To calculate the mass of solid silver sulfide formed, we need to determine the limiting reactant in the reaction between hydrosulfuric acid (H2S) and silver nitrate (AgNO3). The balanced chemical equation for the reaction is:

2H2S(aq) + AgNO3(aq) → Ag2S(s) + 2HNO3(aq)

Using the molar masses of H2S and AgNO3, we can calculate the number of moles of each reactant:

  1. H2S: 5.00 g / 34.08 g/mol = 0.147 mol
  2. AgNO3: 5 g / 169.87 g/mol = 0.0294 mol

The ratio of moles between H2S and AgNO3 in the balanced equation is 2:1. Since there are fewer moles of AgNO3, it is the limiting reactant. Therefore, all the H2S will react with 0.0294 mol of AgNO3 to form 0.0294 mol of Ag2S. To calculate the mass of Ag2S, we multiply the number of moles by its molar mass: 0.0294 mol * 247.8 g/mol = 7.26 g.

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