What volume would be occupied by 88.0 g of gaseous carbon monoxide at STP?

Answer 1

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STP is Standard Temperature and Pressure, so you have to use the formula PV=nrt

Where:

P = Pressure = 1 atm

V = Volume = (unknown)

n = moles of CO

R = gas constant = 0.0821 L*atm / Kmol

T = Temperature = 273K

First do a mass to mole conversion for carbon monoxide (CO)

n = weight/molar mas

weight = 88g (the given mass of CO)

molar mass = 28 (C =12g/mol, O = 16g/mol : 12 + 16 = 28g/mol)

n = 88g / 28g/mol

n = 3.1429 moles

Now solve for Volume:

V = nRT/P

V = (3.1429)(0.0821)(273) / (1)

V = 70.4418 L

The volume occupied by 88.0g of CO gas at STP is 70.4 L