Answer :
Final answer:
The mass of CO₂ formed is 0.2022 grams when 37.1 mL of a 0.124 M Na₂CO₃ solution reacts with a 0.150 M HNO₃ solution, based on stoichiometry and molar mass calculations. However, this result does not match any of the provided options, indicating a possible error.
Explanation:
To calculate the mass of carbon dioxide formed when 37.1 mL of a 0.124 M solution of Na₂CO₃ completely reacts with a 0.150 M solution of HNO₃, we must first write the balanced chemical equation for the reaction:
Na₂CO₃ (aq) + 2 HNO₃ (aq) → 2 NaNO₃ (aq) + CO₂ (g) + H₂O (l)
Next, we need to calculate the moles of Na₂CO₃ using its volume and molarity:
Moles of Na₂CO₃ = Volume × Molarity = 0.0371 L × 0.124 mol/L = 0.004594 moles
From the balanced equation, we can see that there is a 1:1 molar ratio between Na₂CO₃ and CO₂. This means that the moles of Na₂CO₃ are equal to the moles of CO₂ produced.
Finally, to find the mass of CO₂, we use the molar mass of CO₂ (44.01 g/mol):
Mass of CO₂ = Moles × Molar Mass = 0.004594 moles × 44.01 g/mol = 0.2022 grams
The mass of carbon dioxide formed in the reaction, which is closest to 0.2022 grams, is not represented by any of the options given (A) 2.32g, (B) 0.66g, (C) 1.16g, or (D) 3.45g. Therefore, the answer must be recalculated or clarified with more information if there was an error in the given options or the question setup.
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