Question

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.

Answer 1

Answer: 2 moles of water.

Explanation:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given moles:

moles of oxygen = 2.5

moles of $C_3H_8$ = 4.6

According to stoichiometry:

5 moles of oxygen reacts with = 1 mole of propane

Thus 2.5 moles of oxygen will react with = $\frac{1}{5}\times 2.5=0.5$ 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 =$\frac{4}{1}\times 0.5=2moles$ of water

Thus 2 moles of water are produced.