Burning 87.81 grams of hexane will release __ g of water.

__ c6h14 + __ o2 → __ co2 + __ h2o

= 128.772 g of water

2C6H14 + 13 O2 → 6CO2 + 14 H2O

1 mole of Hexane contains 86 g

Therefore;

87.91 grams of hexane will contain

= 87.91 g/86 g

= 1.022 moles

The balanced reaction for the combustion of Hexane is given by

2C6H14 + 13 O2 → 6CO2 + 14 H2O

Therefore; the mole ratio of C6H14 :H2O is

= 2 : 14

= 1: 7

therefore moles of water from 1.022 moles of Hexane will be;

= 1.022 ×7

= 7.154 moles

Mass of water will be;

= 7.154 moles× 18

= 128.772 g of water

The reaction will produce 128.772 g of water

The balanced chemical equation is written as

2C6H14 + 13 O2 → 6CO2 + 14 H2O

Finding the mole ratio of C6H14 :H2O we get

= 2 moles of C6H14 : 14 moles of water H2O

Simplifying it we get 1: 7

In this case moles of water from 1.022 moles of Hexane will be;

moles of water x moles of Hexane

= 1.022 ×7 = 7.154 moles

Now As we know

Moles = Mass / Molar Mass

Mass = Moles x Molar Mass

Mass = 7.154 moles× 18

= 128.772 g of water

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