How many grams of water can be produced by the reaction of 2.16 mol C2H2 with 7.40 mol of oxygen gas?

Question

The heat from an acetylene torch is produced by burning acetylene (C2H2) in oxygen:

2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)

How many grams of water can be produced by the reaction of 2.16 mol C2H2 with 7.40 mol of oxygen gas?

Answer 1

2 C₂H₂ + 5 O₂ → 4 CO₂  + 2 H₂O

 

3              7.5       6            3

2.16           y                      2.16 moles of H₂O (g)

 

2/2.16 = 5/y ------> y = 5.4 moles of oxygen gas; there is a surplus of 2 moles of oxygen that do not react.

x               7.4                    x

 

3/x = 7.5/7.4 ------> x = 2.96 moles of H₂O (g)

 

1 mole of water weighs 18 grams.

After liquefaction, 2.16 times 18 grams can be produced, that is 38.88 grams of water.

Answer 2

The mole ratio of available oxygen to acetylene is 3.4.  The ratio is the equation is 2.5.  So there is excess oxygen, so the limiting reactant is acetylene.

 

So 2.16 moles of acetylene will react, producing 2.16 moles of water (since the coefficients of acetylene and water are the same).

 

2.16 moles of water = 2.16 * (2 + 16) grams, since the atomic mass of hydrogen is 1 and the atomic mass of oxygen is 16).  So the answer is 38.88 g of water.