How many moles of solid Ba(NO3)2 should be added to 300 ml of .2 M Fe(NO3)3 to increase the concentration of NO3- ion to 1 M?
0.300 L x 0.2 M = mols Fe(NO3)3
(NO3^-) = 3 times that = 0.18 mols NO3^-present.
(NO3^-) must be what to make it 1 M. M x L = mols; mols = 1M x 0.300 L = 0.3 mols needed.
We have 0.18 mols so we must add 0.300 - 0.180 = 0.120 mols.
Since Ba(NO3)2 contains 2 mols NO3^-/ mol, then 0.120/2 = 0.060 mols Ba(NO3)2 we must add.
Now we check it to see if we are ok.
mols NO3^- in 300 mL of 0.2 M Fe(NO3)3 = 0.3 L x 0.2 M x (3 mols NO3^-/1 mol Fe(NO3)3 = 0.18 mols present.
Adding 0.06 mols Ba(NO3)2 gives us
0.06 mols Ba(NO3)2 x (2 mols NO3^-/1 mol Ba(NO3)2 = 0.120 mols NO3^-.
0.180 mols NO3^- present at beginning + 0.120 mols NO3^- added with barium nitrate = 0.300 mols NO3^- total.
Molarity = # mols/L = 0.300 mols/0.300 L = 1 M.
Well, let's calculate it in a way that's not "moles"-tifying, shall we?
To increase the concentration of NO3- ions to 1 M, we need to perform a little balancing act worthy of a circus clown. Since Fe(NO3)3 has a 1:3 ratio of Fe to NO3- ions, we need to balance it out with Ba(NO3)2.
Since the concentration of Fe(NO3)3 is 0.2 M, this means it contains 0.2 moles of NO3- ions in 1 liter. Since you have 300 ml, or 0.3 liters, we have:
0.2 moles/L × 0.3 L = 0.06 moles of NO3- ions
To increase this concentration to 1 M, we need to add enough Ba(NO3)2 to produce 0.94 moles of NO3- ions. Since the ratio of NO3- to Ba(NO3)2 is 2:1, we need:
0.94 moles × (1 mole Ba(NO3)2 / 2 moles NO3-) = 0.47 moles of Ba(NO3)2
So, to increase the concentration of NO3- ions to 1 M, you'll need to add approximately 0.47 moles of solid Ba(NO3)2. But remember, this calculation was done by a clown bot, so take it with a grain of laughter!
To calculate the number of moles of solid Ba(NO3)2 needed to increase the concentration of NO3- ion to 1 M, let's follow these steps:
Step 1: Determine the volume of the solution after adding Ba(NO3)2.
Given that 300 ml of Fe(NO3)3 solution is used, the volume of the final solution will be the same if we assume the volume of Ba(NO3)2 solution added is negligible.
Step 2: Calculate the moles of NO3- ions in the desired 1 M solution.
Since the concentration of NO3- ions is given as 1 M and the final volume is the same as the initial volume (300 ml), the number of moles of NO3- ions is:
Moles of NO3- = Concentration × Volume
= 1 M × 0.3 L (converting ml to L)
= 0.3 moles
Step 3: Determine the mole ratio between NO3- ions and Ba(NO3)2.
From the balanced chemical equation, we can see that 1 mole of Ba(NO3)2 produces 2 moles of NO3- ions.
Step 4: Calculate the moles of Ba(NO3)2 needed.
Moles of Ba(NO3)2 = Moles of NO3- ions / Mole ratio
= 0.3 moles / 2
= 0.15 moles
Therefore, you would need to add 0.15 moles of solid Ba(NO3)2 to the 300 ml of 0.2 M Fe(NO3)3 to increase the concentration of NO3- ions to 1 M.
To determine the number of moles of solid Ba(NO3)2 needed to increase the concentration of NO3- ion to 1 M in a solution, we need to first understand the stoichiometry of the reaction.
The balanced equation for the reaction between Ba(NO3)2 and Fe(NO3)3 is:
Ba(NO3)2(aq) + Fe(NO3)3(aq) -> Ba(NO3)2(aq) + Fe(NO3)3(aq)
From this equation, you can see that for every 1 mole of Ba(NO3)2, you also get 2 moles of NO3- ions.
Given that the initial volume of the Fe(NO3)3 solution is 300 ml and its concentration is 0.2 M, we can calculate the initial number of moles of NO3- ions in the solution:
Moles of NO3- = concentration × volume
= 0.2 mol/L × 0.3 L
= 0.06 moles
Now, we need to calculate the additional number of moles of NO3- ions required to increase the concentration to 1 M:
Additional moles of NO3- = (final concentration - initial concentration) × volume
= (1 mol/L - 0.2 mol/L) × 0.3 L
= 0.8 moles
Since every mole of Ba(NO3)2 produces 2 moles of NO3- ions, we divide the additional moles of NO3- by 2 to get the number of moles of Ba(NO3)2 required:
Moles of Ba(NO3)2 = additional moles of NO3- / 2
= 0.8 moles / 2
= 0.4 moles
Therefore, you would need to add 0.4 moles of solid Ba(NO3)2 to the solution to increase the concentration of NO3- ions to 1 M.