Consider the reaction 2H2(g)+O2(g)→2H2O(l) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l ) What is the mass of water, H2O(l) H 2 O ( l ) , produced when 5.70 g g of O2(g) O 2 ( g ) reacts with excess H2(g) H 2 ( g ) ?hen 5.70 g g of O2(g) O 2 ( g ) reacts with excess H2(g) H 2 ( g ) ?

mH₂O = 6.4116 g

Explanation:

Let's write the given reaction:

2H₂ + O₂ > 2H₂O

The problem states that 5.7 g of oxygen reacts with excess hydrogen, hence, the limiting reagent is the oxygen. With this mass of oxygen, we can determine the moles, and then, the moles of water with the mole ratio:

moles = mass / atomic weight        AW of O₂ = 16 g/mol

Replacing we have:

moles O₂ = 5.70 / (16 * 2)

moles O₂ = 0.1781 moles

According to the balanced reaction, 1 mole of Oxygen produces 2 moles of water, so we have a mole ratio 1:2, therefore the moles of water would be twice the moles of oxygen:

moles H₂O = 0.1781 * 2 = 0.3562 moles

Finally the mass of water can be calculated solving for the mass from the expression of moles, and using molecular mass of water:

m = moles * MM          MM H₂O = 18 g/mol

m = 0.3562 * 18

Hope this helps

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