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From your calculations, which reaction experiment had closest to stoichiometric quantities? How many moles of NaHCO₃, and HC₂,H₃,O₂ were present in this reaction?

User EcologyTom
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Final answer:

To determine the reaction experiment closest to stoichiometric quantities, compare the molar ratios of the reactants to the balanced chemical equation. Convert the amounts of NaHCO₃ and HC₂H₃O₂ to moles using their molar masses.

Step-by-step explanation:

In order to determine which reaction experiment had the closest to stoichiometric quantities, you would need to compare the molar ratios of the reactants in each experiment to the balanced chemical equation for the reaction.

For example, if the balanced chemical equation is:

2NaHCO₃ + HC₂H₃O₂ → Na₂C₂H₃O₂ + CO₂ + H₂O

You would need to determine the molar ratio of NaHCO₃ to HC₂H₃O₂ in each experiment. The experiment with the molar ratio closest to 2:1 would have the closest to stoichiometric quantities.

As for the number of moles of NaHCO₃ and HC₂H₃O₂ present in the reaction, you would need to know the amounts of each substance used in the experiment. Once you have the amounts in grams, you can convert them to moles using the molar mass of each substance.

For example, if you have 10 grams of NaHCO₃, you can use the molar mass of NaHCO₃ to convert the grams to moles.

NaHCO₃ molar mass = 23 g/mol + 1 g/mol + 12 g/mol + 16 g/mol + 16 g/mol = 84 g/mol

10 grams NaHCO₃ × (1 mol NaHCO₃ / 84 g NaHCO₃) = 0.119 mol NaHCO₃

Repeat the same calculation for HC₂H₃O₂ to determine the number of moles of HC₂H₃O₂ present in the reaction.

User Paiv
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Final answer:

Without the exact data for the reaction involving NaHCO₃ and HC₂H₃O₂, we can't determine the closest to stoichiometric quantities. The example with NaOH and CH₃CO₂H shows that to be stoichiometric, reactants must be in the molar ratio indicated by the balanced equation.

Step-by-step explanation:

To determine which reaction experiment had the closest to stoichiometric quantities, we would need to compare the mole ratios of the reactants to the mole ratios indicated by the balanced chemical equation. In the case provided, assuming that the reaction in question is the reaction of sodium bicarbonate (NaHCO₃) with acetic acid (HC₂,H₃,O₂), we do not have the exact figures for this reaction. However, we can examine the provided data about another reaction involving acetic acid (CH₃CO₂H) and sodium hydroxide (NaOH) to illustrate how to assess stoichiometry.

The balanced chemical equation for the neutralization reaction is:

NaOH(aq) + CH₃CO₂H(aq) → NaCH₃CO₂(aq) + H₂O(l)

In this equation, the stoichiometry shows a 1:1 molar ratio of NaOH to CH₃CO₂H. According to the data, 5.00 mL of a 1.00 mmol/mL NaOH solution was used, so 5.00 mmol of NaOH were present. Since it reacts with CH₃CO₂H in a 1:1 ratio, we initially have 5.00 mmol of CH₃CO₂H. After the reaction, there's an excess of 4.00 mmol CH₃CO₂H, indicating that there was more CH₃CO₂H than NaOH in the mixture, and therefore the mixture was not stoichiometric.

Stoichiometry is important in ensuring that reactions proceed efficiently with minimal waste of reactants. The experiment with the least excess of any reactant after the reaction would be closest to using stoichiometric quantities.

To determine the number of moles of NaHCO₃ and HC₂H₃O₂ in the student's original reaction, one would need the balanced equation and the volumes and concentrations of the solutions used, which is information not provided.

User CSC
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