Answer:
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
[tex]\because Number\,of\,moles (n) = \frac{mass (m)}{molar\,mass (M)}[/tex]
[tex]\Rightarrow Mass (m) = Number\,of\,moles (n)\times molar\,mass (M)[/tex]
∴ 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 =
[tex]\frac{2.87 g \times 82.03 g}{84 g}= 2.803 g[/tex]
Therefore, the theoretical yield of the reaction = 2.803 g
