Frequently Asked Question

a. The balanced chemical equation is:
2AgNO₂ (aq) + Na₂S (aq) → Ag₂S (s) + 2NaNO₂ (aq)
b. To determine the limiting reactant and calculate the grams of silver sulfide produced, we need to perform the following steps:
1. Convert grams of each reactant to moles using their molar masses:
- Moles of AgNO₂: 35.5 g / (169.87 g/mol) = 0.209 mol
- Moles of Na₂S: 35.5 g / (78.05 g/mol) = 0.455 mol
2. Determine the limiting reactant by comparing the mole ratios of the reactants to the balanced equation. The reactant that produces the least amount of product is the limiting reactant.
- According to the balanced equation, 2 moles of AgNO₂ react with 1 mole of Na₂S. Therefore, 0.209 mol of AgNO₂ would require 0.1045 mol of Na₂S (0.209 mol AgNO₂
(1 mol Na₂S / 2 mol AgNO₂)). Since we have 0.455 mol of Na₂S, it is in excess.
- Therefore, AgNO₂ is the limiting reactant.
3. Calculate the moles of Ag₂S produced using the mole ratio from the balanced equation:
- Moles of Ag₂S: 0.209 mol AgNO₂
(1 mol Ag₂S / 2 mol AgNO₂) = 0.1045 mol Ag₂S
4. Convert moles of Ag₂S to grams using its molar mass:
- Grams of Ag₂S: 0.1045 mol
(247.8 g/mol) = 25.9 g
Therefore, 25.9 grams of silver sulfide will be produced.
c. To determine the amount of silver nitrite remaining, we need to calculate the moles of silver nitrite that reacted. Since silver nitrite is the limiting reactant, all 0.209 mol will react. To find the mass of silver nitrite remaining, subtract the mass of the reacted silver nitrite from the initial mass:
- Grams of AgNO₂ reacted: 0.209 mol
(169.87 g/mol) = 35.5 g
- Grams of AgNO₂ remaining: 35.5 g - 35.5 g = 0 g
Therefore, no silver nitrite will remain at the end of the reaction.