The amount of water (g) produced by combustion of 286 g of propane is
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Combustion of methane :
CH
4
+2O
2
⟶CO
2
+2H
2
O
Given weight of methane =16g
Molecular weight of methane =16g
No. of moles of methane in 16g of it =
mol. wt.
given wt.
=
16
16
=1 mole
From the reaction,
Amount of water produced by the combustion of 1 mole of methane = 2 moles
weight of 1 mole of water =18g
⇒ weight of 2 moles of water =2×18=36g
Since, the given weight of methane is 16 g, hence the water produced by the combustion of 16 g of methane is 36 g.
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