Final answer:
To find the volume of 0.260 M Li2S needed to react with 35.00 ml of 0.315 M AgNO3, we can set up a stoichiometric ratio using the balanced chemical equation and solve for the unknown volume. The correct answer is approximately a. 21.2 ml.
Step-by-step explanation:
To find the volume of 0.260 M Li2S needed to react with 35.00 ml of 0.315 M AgNO3, we need to first determine their stoichiometric ratio using the balanced chemical equation. The balanced equation for the reaction between Li2S and AgNO3 is:
2Li2S + 3AgNO3 -> Ag2S + 6LiNO3
From the equation, we can see that 2 moles of Li2S react with 3 moles of AgNO3. Therefore, we can set up a ratio:
(0.260 M Li2S) / (3 M AgNO3) = (x ml Li2S) / (35.00 ml AgNO3)
Solving for x gives us the volume of Li2S needed, which is approximately 21.2 ml. Therefore, the correct answer is option a. 21.2 ml.
To determine how many milliliters of 0.260 M Li₂S are needed to react with 35.00 ml of 0.315 M AgNO₃, we need to use stoichiometry based on the balanced chemical equation. First, we calculate the number of moles of AgNO₃ we have:
moles AgNO₃ = Molarity AgNO₃ x Volume AgNO₃ (in Liters)
Then, according to the balanced equation, the mole ratio of AgNO₃ to Li₂S will determine the moles of Li₂S needed. Finally, we can work out the volume of Li₂S required using its molarity.