Question

Consider the reaction:   2 H₂ + O₂   → 2 H₂O

What if 3.0 mol H₂ and 2.0 mol O₂ were allowed react. The limiting reactant is ______.

Complete consumption of the limiting reactant would mean the consumption of _____ mol of the other reactant; _____ mol of excess reactant would remain unreacted if the reaction went to completion. The theoretical yield is ____ mol or ____ g of _____.

What if only 2.85 mol of product was obtained? Then, we say that the percent yield is _____%.

Answer 1

In the given reaction, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

We are supplied with 3 mole H2 and 2 mole O2.

2 moles of oxygen require 2*2 = 4 moles hydrogen for its complete reaction.

But we only have 3 moles of hydrogen. Hence, hydrogen in the limiting reagent.

So, 3 moles of hydrogen require 3/2 =1.5 moles of oxygen for complete combustion.

Hence, left over moles of oxygen =2-1.5 = 0.5

Theorerical yield of water :

2 moles hydrogen produce 2 moles of water. Hence 3 moles of hydrogen produce 3 moles of water.

1 mole water weighs : 18 g

Hence, 3 moles water weigh = 3*18 = 54g

If 2.85 moles obtained, percent yield = (2.85/ 3) *100 = 95 %

Hence, the answer in sequence is :

1. Hydrogen
2. 1.5
3. 0.5
4. 3
5. 54
6. 95