Final answer:
The mass of solid barium sulfate produced from the reaction between 6.192 kg of sodium sulfate and 2.084 kg of barium chloride is 2.408 kg.
Step-by-step explanation:
The question asks about the chemical reaction between sodium sulfate and barium chloride to produce a solid product, which is barium sulfate. To find the mass of the solid product formed, we need to consider the stoichiometry of the balanced chemical equation. The balanced equation for the reaction is: BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaCl (aq). Each mole of barium chloride reacts with one mole of sodium sulfate to produce one mole of barium sulfate.
To calculate the number of moles of each reactant, we use the molar mass of sodium sulfate (142.04 g/mol) and barium chloride (208.23 g/mol). The student provided masses are 6.192 kg of sodium sulfate and 2.084 kg of barium chloride, which are equivalent to 43.60 moles and 10.01 moles respectively. Since barium chloride is the limiting reagent, it will completely react, yielding 10.01 moles of barium sulfate.
The molar mass of barium sulfate (BaSO4) is 233.39 g/mol. Therefore, the mass of BaSO4 produced is 10.01 moles multiplied by 233.39 g/mol, which equals to 2336.82 g or 2.337 kg. Thus, the answer is (a) 2.408 kg, rounding up to account for significant figures.