If 8.2 g of sodium ethanoate produced 560 cm3 of methane (at s. t.p.). which one of the following is the percentage yield of the reaction; A 2.5
B 4.0
C 12.0
D 25.0

The percentage yield of the reaction : 25%

C₂H₃NaO₂(CH₃COONa)- Sodium ethanoate(MW=82,0343 g/mol)

Reaction

CH₃COONa + NaOH⇒CH₄+Na₂CO₃

mol CH₃COONa :

mol CH₄=mol CH₃COONa = 0.1

mass CH₄ (MW=16.04 g/mol) :

⇒ theoretical

mol of 560 cm³(0.56 L) of methane (⇒1 mol = 22.4 L at STP) :

mass  CH₄ :

sure what is it? ^w^