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

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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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• Anonymous

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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• Anonymous

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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• 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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