Chemistry: In the stoichiometry laboratory experiment you reacted sodium bicarbonate with acetic acid to form sodium acetate, water and carbon dioxide according to the following equation. nahco3 + ch3cooh → h2o + co2 + nach3co2 calculate the theoretical yield of the reaction if you reacted 2.87 grams of sodium bicarbonate with sufficient acetic acid to produce sodium acetate. group of answer choices

In the stoichiometry laboratory experiment you

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12.01.2022

The theoretical yield of the reaction = 2.803 g

Explanation:

Given reaction: NaHCO₃ + CH₃COOH → H₂O + CO₂ + NaCH₃CO₂

Number of moles of NaHCO₃: n₁ = 1, Number of moles of NaCH₃CO₂: n₂ = 1

Given mass of sodium bicarbonate (NaHCO₃) = 2.87 g

Molar mass of NaHCO₃: M₁ = 84 g/mol, Molar mass of NaCH₃CO₂: M₂ = 82.03 g/mol

$\because Number\,of\,moles (n) = \frac{mass (m)}{molar\,mass (M)}$

$\Rightarrow Mass (m) = Number\,of\,moles (n)\times molar\,mass (M)$

∴ Mass of NaHCO₃ = n₁ × M₁ = 1 mole × 84 g/mol = 84 g

Mass of NaCH₃CO₂ = n₂ × M₂ = 1 mole × 82.03 g/mol = 82.03 g

As in the given reaction, 1 mole NaHCO₃ reacts to give 1 mole NaCH₃CO₂

∴ 84 g NaHCO₃ reacts to give 82.03 g NaCH₃CO₂

⇒ Amount of NaCH₃CO₂ produced when 2.87 g of NaHCO₃ reacts =

$\frac{2.87 g \times 82.03 g}{84 g}= 2.803 g$

Therefore, the theoretical yield of the reaction = 2.803 g

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