Question

g If 1.00 mol of ethane gas and 5.00 mol of oxygen gas react, what is the limiting reactant and how many moles of water are produced from the reaction

Answer 1

Answer:

First, write a balanced equation of the reaction.

A hydrocarbon reacting with oxygen usually gives out carbon dioxide and water (combustion)

2C2H6 + 7O2 ---> 4CO2 + 6H2O

From the equation, we can see the mole ratio of ethane : oxygen is 2:7, meaning 2 moles of ethane reacts with 7 moles of oxygen.

If there's 1 mole of ethane gas, we need y moles of oxygen gas for complete reaction.

$\frac{2}{7} =\frac{1}{y}$

y= 3.5 moles

Since 5 moles of oxygen is more than the required 3.5 moles, we can deduce oxygen gas is in excess, meaning ethane is limiting.

(You can get the same results too if you take y as ethane gas required).

The no. of moles of product produced ALWAYS depend on the no. of moles of the limiting reactant, since they are the ones which reacts completely.

So, from the equation, since the mole ratio of ethane : water = 2:6,

hence, (take the no. of moles of water produced as z),

$\frac{2}{6} =\frac{1}{z} \\z= 3$

Therefore, 3 moles of water is produced.