Challenging Chemistry Equation

Question

How many liters of oxygen gas measured at 22 C and 763 torr are consumed in the complete combustion of 2.55 of dimethyl eter measured at 25 C and 748 torr?

MeOMe 3O2 ----> 2 CO2 3 H2O???? Please help!!

Answer 1

Moles = pV/RT

T = 25 + 273 = 298 K

p = 748 / 760 = 0.984 atm

n = 0.984 x 2.55 / 0.0821 x 298 =0.103

the ratio between MeOMe and O2 is 1 : 3

Moles O2 needed = 0.103 x 3 = 0.309

T = 22 + 273 = 295 K

p = 763/760 = 1.00 atm

V = nRT / p

V = 0.309 x 0.0821 x 295 / 1.00 =7.48 L of O2

Answer 2

When you write a string of chemical symbols interspersed with spaces it is not always obvious that you got them right.....I think this is what you meant:

CH3OCH3 + 3O2 ----> 2CO2 + 3H2O

For convenience, I assume you are talking about 2.55 L of DME.

For a quick answer, convert this quantity to 2.55*273*748/298*760 = 2.299 L at STP.

The molar volume of an ideal gas at STP is 22.414 L, so the amount of DME is 2.299/22.414 = 0.1026 moles

Each mole of DMA requires 3 moles of O2 for a total of 0.3077moles O2. At STP, this is 0.3077*22.414 = 6.898 L. Correcting to 22°C and 763 torr, we get:

6.898*295*760/273*763 = 7.42 L (3 significant figures) of O2

Answer 3

I presume 2.55 L of MeOMe

Moles = pV/RT

T = 25 + 273 = 298 K

p = 748 / 760 = 0.984 atm

n = 0.984 x 2.55 / 0.0821 x 298 =0.103

the ratio between MeOMe and O2 is 1 : 3

Moles O2 needed = 0.103 x 3 = 0.309

T = 22 + 273 = 295 K

p = 763/760 = 1.00 atm

V = nRT / p

V = 0.309 x 0.0821 x 295 / 1.00 =7.48 L of O2