The combustion of propane (c3h8) produces co2 and h2o: c3h8(g) 5o2(g)→3co2(g) 4h2o(g) the reaction of 2.5 mol of o2 with 4.6 mol of c3h8 will produce mol of h2o.

2 moles of water.

Explanation:

Given moles:

moles of oxygen = 2.5

moles of = 4.6

According to stoichiometry:

5 moles of oxygen reacts with = 1 mole of propane

Thus 2.5 moles of oxygen will react with = moles of oxygen

Thus propane is the limiting reagent as it limits the formation of products and oxygen is the excess reagent.

As 1 mole of propane produce = 4 moles of water

0.5 mole of propane will produce = of water

Thus 2 moles of water are produced.

anaphase.

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