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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.