Based on the tiles shown, how many moles of oxygen gas (o2) are needed to produce 36.04 grams of water (h2o) when reacting with excess ethane in the equation 2c2h6 7o2 → 5co2 6h2o?

Respuesta :

We calculate for the number of moles of water given its mass by dividing the given mass by the molar mass. 
                               n water = (36.04 g) / (18 g/mol) 
                                n water = 2 mols
From the given balanced equation, every 6 moles of water produced will require 7 moles of oxygen.
                                n oxygen = (2 mols H2O) x (7 moles O2 / 6 moles H2O)
                                n oxygen  = 2.33 mols O2

The number moles of oxygen that produced 36.04 grams of water is 2.33 moles of oxygen.

The equation of the reaction is;

2C2H6 + 7O2 → 5CO2 + 6H2O  

We were told in the question that ethane is in excess hence oxygen is the limiting reactant.

Number of moles of water = 36.04 g/18g/mol = 2 moles

If  7 moles of oxygen produces 6 moles of water

x moles of oxygen produces   2 moles of water

x =  7 moles ×  2 moles/6 moles

x = 2.33 moles of oxygen

Mass of oxygen =  2.33 moles of oxygen ×  32 g/mol = 74.56 g

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