# Forty grams of CH4 gas at STP occupies what volume?

### 7 Answers

- 1 decade agoFavorite Answer
PV = nRT

P= 1 atm

R=.0821

T=273 K

n=weight / mol.mass = 40/16

therefore, V = nRT/P

V = (40*0.0821*273)/(16*1)

= 56 Litres

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- Anonymous1 decade ago
The molar mass of CH4 is 16 grams per mole. One mole of any gas at STP occupies 22.4 Liters. Since you have 40 grams you divide that by the 16 grams and get the number of Liters of gas. 40/16 = 20/8 = 10/4 = 2.5 Liters.

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- Anonymous1 decade ago
This question can be answered using the ideal gas equation PV=nRT

n=no. of moles=mass of the substance/molar mass=40/12+4=40/16=2.5 moles. V=nRT/P=2.5 x 0.0821 x 273.15K/0.987atm=56.8 liters. R=gas constant

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- Trevor HLv 71 decade ago
Molar mass CH4 = 12 + 4 =16

No. of moles in 40g = 40/16 = 2.5moles

PV = nRT

101.3 V = 2.5*8.31*273

V = 56.0 litres

Chech 1mol at STP = 22.4 litres

2.5 moles at STP = 2.5*22.4 = 56.0litres

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- Andrew LLv 41 decade ago
If you are using the ideal gas law, use pV=nRT where p=pressure, V=volume, n=number of moles, R=universal gas constant, T=temperature.

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- 1 decade ago
CH4=16g/mole ; 1mole=22.4L/mole@stp

40g/16g/molex22.4L/mole=56.0L@stp

Source(s): chem major- Login to reply the answers

- 4 years ago
40(mass) / 14.01 (molar mass) = 2.855 moles 2.855 x 22.4 = 63.95 L

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