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

Question

asked Feb 10, 2021 161k views

1 Answer

Narek Tootikian · Answer 1 · 2021-02-16T12:32:40+0000

The percentage yield of the reaction : 25%

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 :

$\tt 0.025* 16.04=0.401~g$

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