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
<em>- Dioxygen is the limiting reagent. </em>
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).
<em>We can setup a proportion;</em>
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?
<em><u>0.196 mol of H2O x 18 g/mol of H2O = 3.53 grams of H2O </u></em>
It increases. B.
Close to the nucleus are high energy levels, because of the pull of the nucleus. But farther away from the nucleus, the electrons there are of low energy level.
To answer this, we use Raoult's Law where the partial pressure is equal to the product of the fraction of the gas in the mixture and its total pressure. Also, we use Dalton's Law of Partial Pressure where the total pressure is equal to the sum of the partial pressure of the gases in the mixture.
Ptotal = 73.44 + 128.52 + 2.04 = 204 atm
201.96 = x (204)
x = 0.99
