How many millimeters of .230 M Na2S are needed to react with 30.00 mL of .513 M AgNO3, according to the following balanced equation? Na2S (aq) + 2AgNO3 (aq) ---> 2NaNO3 (aq) + Ag2S (s)

Step 1: Draft the chemical equation that is balanced.

To solve this problem, we first need to determine the number of moles of AgNO3 in 30.00 mL of 0.513 M solution. Then, using the stoichiometry of the balanced equation, we can find the number of moles of Na2S required to react with the moles of AgNO3. Finally, we convert the moles of Na2S to millimeters using its molarity.

1. Moles of AgNO3: Molarity (M) = moles / volume (L) Moles = Molarity × volume (L) Moles of AgNO3 = 0.513 mol/L × 0.030 L = 0.01539 moles

2. According to the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2S. So, 0.01539 moles of AgNO3 will react with 0.007695 moles of Na2S.

3. Now, we use the molarity of Na2S to find the volume needed: Molarity (M) = moles / volume (L) Volume (L) = moles / Molarity Volume (L) = 0.007695 moles / 0.230 mol/L = 0.03346 L

   Converting liters to millimeters: 0.03346 L × 1000 mL/L = 33.46 mL

Therefore, 33.46 millimeters of 0.230 M Na2S are needed to react with 30.00 mL of 0.513 M AgNO3.