# Reaction Formula

1. Oct 13, 2004

### Mathman23

Hi

I have this reaction formula

$CH_{4} + 2O_{2} \rightarrow 2H_{2} O + CO_{2}$

CH4 and O2 are contained in a canister which has a volume V = 500 Liters.

The mass of CH4 is 100 grams and O2 has a mass of 100 grams.

I have two questions:

First if I want to calculate the mass of the generated $CO_2$.

Can't this be done by the following calculation

$n(CH_4) = n(CO_2) \ \rightarrow \ n(CO_2) = \frac{m(CH_4)}{M(CH_4)} = ??$

And then multiply it with $M(CO_2)$ ??

Second: if the temperature in the canister is messured to be 105 degrees celcius at the end of the above reaction. How do I calculate the partial pressure of $CH_{4}$ at this temperature ??

Sincerely Fred
Denmark

Last edited: Oct 14, 2004
2. Oct 14, 2004

### wolfson_1123

CH4+ 202 --> 2H2O + CO2

CH4 = 100g

2O2 = 100g

Mr of substances (one mole weighs):

CH4 = 16.043g

O2 = 31.999 so 2O2 = 63.998

H2O = 18.015 so 2H2O = 63.030

CO2 = 44.010.

Now calculate number of moles:

Number of moles = Mr / w (w = mass of substance)

CH4 = 100 / 16.043 = 6.23 mol.

02 = 100 / 69.998 = 1.43 mole.

Mole of CO2 = 6.23 x 1/1.43 = 4.40 mol

Mass = 4.4 x 44.010 = 193.644g.

Part 2:

Temperature = 105 + 273.15 = 378.15K

Volume = 500 dm^3

pO2 = 6.23(0.0821 (375.15)) / 500 = 0.68 atm.

Wolfson.

3. Oct 14, 2004

### Mathman23

Hi

How is possible that the partial pressure of O2 molecules at the end of the reaction are the same as the p.pressure of CH4 molecules?

Sincerely
Fred

4. Oct 16, 2004

### garytse86

from the mole fraction of CH4

5. Oct 17, 2004

Sincerely

/Fred