1. The volume occupied by 3.60 g of H₂ gas at stp is 40.32 L.
2. The law used in this case is Avogadro's law.
We'll begin by writing calculating the number of mole in 3.60 g of H₂. This can be obtained as follow:
Mass of H₂ = 3.60 g
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mole of H₂ =?
Mole = mass / molar mass
Mole of H₂ = 3.6 / 2
Mole of H₂ = 1.8 moles
From Avogadro's law (which states that equal volumes of all gaes contains the same number of molecules at constant temperature and pressure), we understood that:
1 mole of a gas occupies 22.4 L at stp.
With the above information in mind, we can obtain the volume occupied by 3.60 g (i.e 1.8 moles) of H₂. This illustrated below:
1 mole = 22.4 L
Therefore,
1.8 moles = 1.8 × 22.4
1.8 moles = 40.32 L
Therefore, 3.60 g (i.e 1.8 moles) of H₂ occupies 40.32 L at stp
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