i was practicing some advanced level questions and i came across this one

On heating with soda lime, sodium ethanoate decomposes in accordance with the following eqn

CH3COONa + NaOH ----> CH4 + Na2CO3

iF 8.2g of sodium ethanoate produced 560 cm^3 of methane (at stp (standard temp. and pressure) what is the percentage yield?

they have the answer as

25.0

To find the percentage yield, we need to compare the actual yield to the theoretical yield. Here's how you can solve this problem step-by-step:

Step 1: Calculate the molar mass of sodium ethanoate (CH3COONa):
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol
Na: 22.99 g/mol

Molar mass of CH3COONa = (12.01 * 2) + (1.01 * 3) + 16.00 + 22.99 = 82.03 g/mol

Step 2: Determine the number of moles of sodium ethanoate (CH3COONa) used:
Mass of sodium ethanoate = 8.2 g
Number of moles = mass / molar mass
Number of moles of CH3COONa = 8.2 g / 82.03 g/mol = 0.099909 mol (approximately 0.1 mol)

Step 3: Based on the balanced equation, determine the stoichiometry between sodium ethanoate (CH3COONa) and methane (CH4):
From the balanced equation: CH3COONa + NaOH → CH4 + Na2CO3
1 mole of CH3COONa produces 1 mole of CH4

Step 4: Calculate the theoretical yield of methane (CH4):
The stoichiometry tells us that 0.1 mol of CH3COONa will produce 0.1 mol of CH4.

Step 5: Use the ideal gas equation to calculate the volume of 0.1 mol of methane at STP (Standard Temperature and Pressure):
1 mole of any gas occupies 22.4 L at STP.

Volume of 0.1 mol of CH4 = (0.1 mol) * (22.4 L/mol) = 2.24 L

Step 6: Compare the given actual yield (560 cm^3) to the theoretical yield (2.24 L):
To compare the values, we need to convert cm^3 to L. 1 L = 1000 cm^3.

Actual yield = 560 cm^3 / 1000 cm^3/L = 0.56 L

Step 7: Calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) * 100%
Percentage yield = (0.56 L / 2.24 L) * 100% = 25%

Therefore, the percentage yield is 25%.