Answer :
The balanced equation for the given mixture is the following.
BaCl2 + Na2CO3 → BaCO3(s) + 2 NaCl
now, calculate the no of moles of each BaCl2 and Na2CO3
(65 mL) x (0.3 M BaCl2) = 0.0195 mol BaCl2
(60 mL) x (0.35 M Na2CO3) = 0.021 mol Na2CO3
0.0195moles of BaCl2 is reactin completely with 0.021moles of Na2CO3, but we have more Na2CO3 present than BaCl2, so Na2CO3 is in excess and BaCl2 is the limiting reactant.
Therefore,
(0.0195 mol BaCl2) x (1 mol BaCO3 / 1 mol BaCl2) x (197.3359 g/mol) = 3.85 g BaCO3.
197.3359 g/mol is the molar mass of the BaCO3.
A precipitate is a solid formed in a chemical reaction other than one of the reactants. This can occur when solutions containing ionic compounds are mixed to form an insoluble product. The identity of precipitates can often be determined by examining solubility rules. Mass is preserved during the precipitation reaction. No matter what changes occur, the total mass of matter contained remains the same. The mass of the product is the same as the mass of the reactants in a chemical reaction.
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