If 1.00 mole of a gas has a mass of 18 grams, and you have 15.0 grams of the gas, what would its volume be at STP?
Molar mass of the gas = 18 g per mol
The mass of gas available = 15 g
Therefore, the number of moles present is given by,
Number of moles = mass / molar mass
Number of moles = 15 g/ 18 g per mol
= 5/6 mol
= 0.83333 mol.
STP - Standard Temperature & Pressure
This is 0 C and 1 bar according to IUPAC definition. At these conditions, the volume of 1 mol of ideal gas is 22.71 L/mol.
If we assume that the given gas is an ideal, volume of 1 mol at STP is 22.71 L/mol. But according to Avogadro Law volume, V is,
V = kn Where k is constant and n is the number of moles.
If the volume of 15 g of gas is V, then,
22.71 = k x 1
V = k x (5/6) (since it has only 5/6 moles)
V = 22.71 x (5/6) = 18.925 L
Therefore, volume of 15 g of the gas is 18.925 L.