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


Sagot :

Answer:

[tex]55[/tex] [tex]grames[/tex] [tex]02\times\left(1\ mol\ 02\div32\ g\ 02\right)\times\left(2\ mol\ H20\div1\ mol\ 02\right)\times\left(22.4\ L\div1\ mol\ H20\right)=77\ L\ HTO\left(g\right)[/tex]

Explanation:

It's about 62 milliliters in liquid form.

I hope this helps you

:)

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