# Question 1dded

Feb 23, 2015

You're dealing with the combustion of propane. The balanced chemical equation looks like this

${C}_{3} {H}_{8} + 5 {O}_{2} \to 3 C {O}_{2} + 4 {H}_{2} O$

Now, the tool you have at your disposal when doing limiting reagent problems (or any stoichiometry problem) is the mole ratio. The balanced chemical equation gives you the proportion in which the reactans must mix; in this case, 1 mole of propane needs 5 moles of oxygen.

Calculate how many moles of each reactant you get by using their molar masses

$\text{14.8 g propane" * ("1 mole")/("44.0 g") = "0.336 moles propane}$

$\text{3.44 g oxygen" * ("1 mole")/("32.0 g") = "0.108 moles oxygen}$

It's obvious that oxygen will act as a limiting reagent, since the mole ratio would have required

$\text{0.336 moles propane" * ("5 moles oxygen")/("1 mole propane") = "1.68 moles oxygen}$

The number of moles of oxygen you have doesn't even come close to this. This means that oxygen will determine how much propane reacts

$\text{0.108 moles oxygen" * ("1 mole propane")/("5 moles oxygen") = "0.0216 moles propane}$

To determine the mass of carbon dioxide produced use the mole ratio $C {O}_{2}$ has with ${C}_{3} {H}_{8}$

$\text{0.0216 moles propane" * ("3 moles CO"_2)/("1 mole propane") = "0.0648 moles}$ $C {O}_{2}$

$\text{0.0648 moles CO"_2 * ("44.0 g")/("1 mole") = "2.85 g}$ $C {O}_{2}$ produced.

Since you only used $\text{0.0216 moles}$ of propane, you'll have an excess of

n_("excess") = 0.336 - 0.0216 = "0.314 moles"#

The mass of excess propane will be

$\text{0.314 moles" * ("44.0 g")/("1 mole") = "13.8 g}$

SIDE NOTE The mass of oxygen is so small that the limiting reagent was bound to be oxygen; the problem would have been a little more interesting (but not much) if you had 34.4 g of oxygen instead of 3.44...