Answer :
Final answer:
The information provided does not match the reaction described in the question, hence it's not possible to calculate the theoretical mass of the precipitate formed from Rubidium Carbonate and Barium Acetate without the correct formula and stoichiometry.
Explanation:
The question asks for the theoretical mass of precipitate formed from a reaction between Rubidium Carbonate and Barium Acetate. We can use stoichiometry to solve this problem, but we need the balanced chemical equation for the reaction of these two compounds. However, the provided information does not match the chemicals mentioned in the question, thus it is not possible to perform an accurate calculation without the correct formula and stoichiometry of the precipitate formed (likely Barium Carbonate).
The conversion factors given for Rb and MgCl2 are not relevant to the formation of a precipitate from Rubidium Carbonate and Barium Acetate, which makes it impossible to provide an accurate response to the question as it stands. One would need the correct molar mass of Barium Carbonate and stoichiometry of the reaction to determine the theoretical yield of the precipitate.
Given that the volume and molarity of barium acetate are 38.2 ml and 5.0 M, respectively. We need to find the theoretical mass in grams of the precipitate. Let's first write the balanced chemical equation for the reaction taking place: Rubidium Carbonate + Barium Acetate → Barium Carbonate + Rubidium AcetateRb2CO3(aq) + Ba(C2H3O2)2(aq) → BaCO3(s) + 2 RbC2H3O2(aq).
We can see that 1 mole of barium acetate reacts with 1 mole of barium carbonate. Hence, the molar ratio of barium acetate and barium carbonate is 1:1.Using the molarity and volume of barium acetate, we can find the moles of barium acetate as: Moles of barium acetate = Molarity × Volume in litres= 5.0 mol/L × (38.2/1000) L= 0.191 moles. Now, from the balanced chemical equation, we can see that 1 mole of barium carbonate is formed from 1 mole of barium acetate.
Therefore, the number of moles of barium carbonate formed will also be 0.191 moles. Now, let's calculate the mass of barium carbonate using its molar mass. Molar mass of BaCO3= (1 × atomic mass of Ba) + (1 × atomic mass of C) + (3 × atomic mass of O)= (1 × 137.33 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)= 197.33 g/mol. Theoretical mass of BaCO3= Number of moles of BaCO3 × Molar mass of BaCO3= 0.191 mol × 197.33 g/mol= 37.7 g. Therefore, the theoretical mass of the precipitate is 37.7 g (approx) when only the volume and molarity of the barium acetate are taken into account. Note: In order to find the limiting reagent and the actual mass of the precipitate formed, we need to consider the volume and molarity of both the reactants.
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