Suppose a chemist mixes 13.8g of ethane and 45.8g of dioxygen. Calculate the theoretical yield of water.

Question

Suppose a chemist mixes 13.8g of Ethane and 45.8g of dioxygen. calculate the theoretical yield of water

Answer 1

1) First, write out the balanced equation:

 

2C₂H₆ + 7O₂ --> 4CO₂ + 6H₂O

 

2) Second, you need to determine the limiting reagents:

 

13.8 grams of C₂H₆/30.06 g/mol of C₂H₆ = 0.46 mol

45.8 grams of O₂/32 g/mol of O₂ = 1.4 mol

 

0.46 mol of C₂H₆/2 = 0.23

1.4 mol of O₂/7 = 0.20

 

Dioxygen is the limiting reagent.

 

3) Find the theoretical molar yield of water:

 

There is a 7 mol of O₂ to 6 mol of H₂O ratio (according to the balanced equation).

 

So, we can setup a proportion in which:

 

7/6 = 0.23/x

0.196 = x

 

So, 0.196 mol of water will be produced.

 

4) Find how many grams of water will be produced in the reaction:

 

0.196 mol of H₂O x 18 g/mol of H₂O = 3.53 grams of H₂O