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# How to calculate the molar gas volume of methane at STP? (chemistry homework)?

Please help me figure this out. If possible explain how you got the answer? This is the exact wording of the problem:

A sample of sodium acetate was mixed with excess sodium hydroxide and heated in a test tube to decompose it completely according to the following equation:

NaCH3COO(s) + NaOH(s) -----> Na2CO3(s) + CH4(g)

The weight loss from the test tube was 0.2043g. The evolved methane gas (CH4) was collected over water at 27.0[degree symbol] and 747.0mmHg resulting in a volume of 321mL of gas collected.

Calculate the molar gas volume of methane at STP.

### 2 Answers

- Lexi RLv 710 years agoFavorite Answer
moles of CH4 = mass lost from test tube, as it is the only gas evolved

moles = mass / molar mass

= 0.2043 g / 16.042 g/mol

= 0.012735 mol CH4

Now, because you have collected the gas over H2O there will be some H2O vapour in the sample contributing to the pressure. So you need to look up the vapour pressure of H2O at 27 deg C, then

partial pressure CH4 = total pressure - vapour pressure H2O at 27 deg

http://intro.chem.okstate.edu/1515sp01/database/vp...

partial pressure CH4 = 747.0 mmHg - 26.7 mmHg

= 720.3 mmHg

Now, that we know the partial pressure of the CH4, we can use the use the combined gas equation to determine the volume of this sample at STP (0 deg Celcius and 760 mmHg).

P1V1 /T1 = P2V2 / T2

P1 = 720.3 mmHg

V1 = 321 ml

T1 = 27.0 deg C = 300.15 K

P2 = 760.0 mmHg

V2 = ? ml

T2 = 273.15 K

V2 = P1V1T2 /T1P2

= 720.3 mmHg x 321 ml x 273.15 K / (300.15 K x 760.0 mmHg

= 276.86 ml

= 0.27686 L

So now we know that 0.012735 mol of CH4 will occupy a volume of 0.27686 L at STP

Molar volume = Litres / moles

= 0.27686 L / 0.012735 mol

= 21.7 L/mol

The actual molar volume of an ideal gas should be 22.4 L/mol