The percentage yield of the reaction : 25%
Further explanation
C₂H₃NaO₂(CH₃COONa)- Sodium ethanoate(MW=82,0343 g/mol)
Reaction
CH₃COONa + NaOH⇒CH₄+Na₂CO₃
mol CH₃COONa :
[tex]\tt \dfrac{8.2}{82.0343}=0.1[/tex]
mol CH₄=mol CH₃COONa = 0.1
mass CH₄ (MW=16.04 g/mol) :
[tex]\tt 0.1\times 16.04=1.604~g[/tex]⇒ theoretical
mol of 560 cm³(0.56 L) of methane (⇒1 mol = 22.4 L at STP) :
[tex]\tt \dfrac{0.56}{22.4}=0.025~mol[/tex]
mass CH₄ :
[tex]\tt 0.025\times 16.04=0.401~g[/tex]
[tex]\tt \%yield=\dfrac{0.401}{1.604}\times 100\%=25\%[/tex]
