Question

How many liters of water can be made from 55 grams of oxygen gas and an excess of

hydrogen at STP?

Answer 1

1 mole of gas at STP = 22.4 L

Reaction:

2H₂(g) + O₂(g) → 2H₂O(l)

mole oxgen as a limiting reactant

mole oxygen: mass : molar mass = 55 g : 32 = 1.719

mole water H₂O based on mole oxygen as a limiting reactant

mole H₂O from reaction coefficient: 2/1 x mole O₂ = 2/1 x 1.719 = 3.438

liters H₂O at STP = 3.438 x 22.4 L = 77.0112 L

Answer 2

`55`
`grames`
`02*\left(1\ mol\ 02/32\ g\ 02\right)*\left(2\ mol\ H20/1\ mol\ 02\right)*\left(22.4\ L/1\ mol\ H20\right)=77\ L\ HTO\left(g\right)`

Step-by-step explanation:

It's about 62 milliliters in liquid form.

I hope this helps you

:)