98 ! must show your work!
2c₂h₆ (g) + 7o₂ (g) --> 4co₂ (g) + 6h₂o (g)
you mix 16.5 grams of ethane (c₂h₆) with 17 grams of oxygen gas (o₂).
a) which gas is the limiting reactant reactant?
b) how many grams of water should be produced?
c) how many grams of the excess reactant is left over?
d) if only 6.6 grams of water are produced, what is the percent yield?

D) If only 6.6 grams of water are produced, what is the percent yield?

Write out the balanced equation:

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

Then, determine the limiting reagents:

16.8 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.  

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).

We can setup a proportion;

7/6 = 0.23/x

0.196 = x

So, 0.196 mol of water will be produced.

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

dynamic equilibrium is the answer.