# What is the mass of BaCl2 contained in 250 mL of solution since after the addition of 40.00 ml of AgNO3 0.1020M to 25.00 mL of the BACl2 solution, the excess solution of AgNO3 consumed 15.00 ml of NH4SCN 0.09800 M

## Well, this sounds like a chemistry problem. Let's see if we can solve it together, shall we?

To find the mass of BaCl2, we need to use stoichiometry. But before we do that, let's analyze the reaction first. When AgNO3 reacts with BaCl2, it forms a precipitate of AgCl. This means that AgNO3 is the limiting reagent in this reaction.

Now, let's use some humor to lighten up the situation. Why did the chemist break up with their significant other? Because there was no "chemistry" between them! Okay, okay, I'll stick to explaining the problem.

Since AgNO3 is the limiting reagent, it means that every AgNO3 molecule reacts with a BaCl2 molecule in a 1:1 ratio. So, the moles of BaCl2 can be calculated by using the volume and molarity of AgNO3 used.

The moles of AgNO3 used can be calculated using the formula: moles = volume (in L) x concentration (in mol/L). In this case, the volume is 40.00 mL, which is equal to 0.04000 L, and the concentration is 0.1020 M.

So, the moles of AgNO3 used would be 0.04000 L x 0.1020 mol/L = 0.00408 mol.

And since the reaction between AgNO3 and BaCl2 is in a 1:1 ratio, the moles of BaCl2 is also 0.00408 mol.

Now, we need to calculate the mass of BaCl2. First, we need to find the molar mass of BaCl2. The molar mass of Ba is 137.33 g/mol, and the molar mass of Cl is 35.45 g/mol. So, the molar mass of BaCl2 would be 137.33 g/mol + (2 x 35.45 g/mol) = 208.23 g/mol.

To find the mass of BaCl2, we multiply the moles of BaCl2 by the molar mass:

Mass = moles x molar mass = 0.00408 mol x 208.23 g/mol ≈ 0.848 g.

Therefore, the mass of BaCl2 contained in 250 mL of solution would be approximately 0.848 g.

I hope that helps! If you have any more chemistry questions or if you need a good joke, feel free to ask!

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## To find the mass of BaCl2 contained in the solution, we need to calculate the amount of BaCl2 before and after the reaction.

1. Calculate the amount of AgNO3 used:
C1V1 = C2V2
(0.1020 M)(40.00 mL) = C2(15.00 mL)
C2 = (0.1020 M)(40.00 mL) / (15.00 mL)
= 0.272 M

2. Calculate the moles of AgNO3 used:
moles of AgNO3 = C2V2
moles of AgNO3 = (0.272 M)(15.00 mL)
= 0.00408 moles

3. Determine the stoichiometry between AgNO3 and BaCl2:
Balanced equation: 2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2
From the equation, we can see that the stoichiometric ratio is 2:1 between AgNO3 and BaCl2.

4. Calculate the moles of BaCl2:
moles of BaCl2 = moles of AgNO3 / 2
moles of BaCl2 = 0.00408 moles / 2
= 0.00204 moles

5. Determine the molar mass of BaCl2:
Molar mass of BaCl2 = (1 × atomic mass of Ba) + (2 × atomic mass of Cl)
= (1 × 137.33 g/mol) + (2 × 35.45 g/mol)
= 137.33 g/mol + 70.90 g/mol
= 208.23 g/mol

6. Calculate the mass of BaCl2:
mass of BaCl2 = moles of BaCl2 × molar mass of BaCl2
= 0.00204 moles × 208.23 g/mol
= 0.4236 grams

Therefore, the mass of BaCl2 contained in the 250 mL solution is approximately 0.4236 grams.

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## To calculate the mass of BaCl2 contained in the solution, we need to first determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

In this case, we have two reactants: AgNO3 and NH4SCN. The reaction between AgNO3 and NH4SCN forms a precipitate of AgSCN. The balanced equation for this reaction is:

AgNO3 + NH4SCN → AgSCN + NH4NO3

We are given the volume and concentration of AgNO3 (40.00 mL, 0.1020 M) and the volume and concentration of NH4SCN (15.00 mL, 0.09800 M). However, we also need to know the stoichiometry of the reaction to determine the moles of each reactant.

From the balanced equation, we can see that the molar ratio between AgNO3 and NH4SCN is 1:1. This means that it takes one mole of AgNO3 to react with one mole of NH4SCN.

To calculate the moles of AgNO3, we use the formula:

moles AgNO3 = volume (L) x concentration (M)

First, convert the volume of AgNO3 from mL to L:

40.00 mL = 40.00 mL x (1 L / 1000 mL) = 0.04000 L

Now, calculate the moles of AgNO3:

moles AgNO3 = 0.04000 L x 0.1020 M = 0.00408 moles AgNO3

Since the stoichiometry of the reaction is 1:1, this means that there are also 0.00408 moles of NH4SCN.

Now, we need to determine the amount of BaCl2 that is required to react with the AgNO3. The balanced equation for the reaction between BaCl2 and AgNO3 is:

BaCl2 + 2AgNO3 → 2AgCl + Ba(NO3)2

From the balanced equation, we can see that the molar ratio between BaCl2 and AgNO3 is 1:2. This means that it takes one mole of BaCl2 to react with two moles of AgNO3.

Since we have 0.00408 moles of AgNO3, we can calculate the moles of BaCl2 that are required:

moles BaCl2 = 0.00408 moles AgNO3 x (1 mole BaCl2 / 2 moles AgNO3) = 0.00204 moles BaCl2

Finally, we need to calculate the mass of BaCl2 using its molar mass. The molar mass of BaCl2 is the sum of the atomic masses of barium (Ba) and chlorine (Cl).

The atomic mass of barium (Ba) is 137.33 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol. Therefore, the molar mass of BaCl2 is:

molar mass BaCl2 = (1 x atomic mass Ba) + (2 x atomic mass Cl)
= (1 x 137.33 g/mol) + (2 x 35.45 g/mol)
= 137.33 g/mol + 70.90 g/mol
= 208.23 g/mol

Finally, calculate the mass of BaCl2:

mass BaCl2 = moles BaCl2 x molar mass BaCl2
= 0.00204 moles x 208.23 g/mol
= 0.423 g BaCl2

Therefore, the mass of BaCl2 contained in the 250 mL solution is 0.423 grams.