How many grams are in 6.8L of Oxygen gas (O2) at STP

9.7 g

Explanation:

From the question,

Note: The molar volume of all gas at stp is 22.4 dm³ or 22.4 L

1 mol of  oxygen gas (O₂) at stp = 22.4 dm³

X mole of oxygen gas (O₂) at stp = 6.8 L

X = (1 mol×6.8 L)/22.4 L

X = 0.3036 mol.

But,

Number of mole (n) = mass (m)/molar mass (m')

n = m/m'

m = n×m' Equation 2

Where n = 0.3036 mol, m' = 32 g/mol

Substitute into equation 2

m = 0.3036×32

m = 9.7 g

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