Question

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

Answer 1

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 :

`\tt (8.2)/(82.0343)=0.1`

mol CH₄=mol CH₃COONa = 0.1

mass CH₄ (MW=16.04 g/mol) :

`\tt 0.1* 16.04=1.604~g`⇒ theoretical

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

`\tt (0.56)/(22.4)=0.025~mol`

mass CH₄ :

`\tt 0.025* 16.04=0.401~g`

`\tt \%yield=(0.401)/(1.604)* 100\%=25\%`