Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Round your answer to significant digits.

Final Molarity of acetate ion = 0.788 M

Explanation:

We first assume the reaction goes to completion and that the volume.of the reaction doesn't change.

The balanced stoichiometric equation is

Ba(CH₃COO)₂ + (NH₄)₂SO₄ → BaSO₄ + 2NH₄(CH₃COO)

First of, we need to check which reactant is in excess and which one is completely used up in the reaction. To do this, we fimd the number moles of each reactant at the start of the reaction.

For Ba(CH₃COO)₂

Number of moles = (Mass)/(Molar Mass)

Mass of Barium acetate = 15.1 g

Molar mass of Barium acetate = 255.43 g/mol

Number of moles = (15.1/255.43) = 0.0591 moles.

For (NH₄)₂SO₄,

Number of moles = (Concentration in mol/L) × (Volume in L)

Concentration of Ammonium surface in mol/L = 0.40 M

Volume in L = (150/1000) = 0.150 L

Number of moles = 0.40 × 0.150 = 0.0600 moles

From the stoichiometric balance of the reaction,

1 mole of Ba(CH₃COO)₂ reacts with 1 mole of (NH₄)₂SO₄

Hence, it is evident that it is Ba(CH₃COO)₂ that is the limiting reagent; the chemical specie that is used up in the process of the reaction and it determines the amount of other reactants required & products formed.

1 mole of Ba(CH₃COO)₂ gives 2 moles of NH₄(CH₃COO)

0.0591 moles of Ba(CH₃COO)₂ will give 2 × 0.0591 moles of NH₄(CH₃COO); 0.1182 moles of NH₄(CH₃COO).

The molarity of NH₄(CH₃COO) is then given as (number of moles) ÷ (Vol in L)

Number of moles = 0.1182 moles

Volume of the solution in L = 0.150 L

Molarity of NH₄(CH₃COO) = (0.01182/0.15)

= 0.788 M

Note that 1 mole of NH₄(CH₃COO) contains 1 mole of acetate ion,

Hence, 0.788 M of NH₄(CH₃COO) also contains 0.788 M of acetate ion.

Hope this Helps!!!

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.

1.solid

2.carbon dioxide

3.helium