Upon the combustion of propane, C3H8, 150. g of propane was reacted with excess oxygen gas.
A. Determine the limiting reactant if the water is the product of interest.
B. Determine the number of grams of water produced.
C. Calculate the number of grams of water obtained in the laboratory given that the percent yield was 68.7%
A. 1 mole of C3H3, there are 5 moles O2
150.gC3H8 (1moleC3H8 / 39.1gC3H8) = 3.84moles C3H8
MMO2 = 32gO2=1moleO2
3.84x5 = 19.2moles O2
Limiting reactant: C3H8
B total product is 71.1g. 71.1 / 2 = 35.6gH2O
C. Not sure. I know the formula is actual yield / theoretical yield x 100 = percent yield.
I do not think I'm doing this problem right. what am I doing wrong? Could you please correct me and explain? I do not understand percent yields, but maybe if I understand the first parts better, I would. Thanks!
excess oxygen gas makes the limiting reactant propane, period.
a. where did you get the 5 moles of O2, what is it matter, you have an excess of O2.
b. In 150g propane, you have 3.84 moles propane, and in that propane, you have 4*3.84 moles H2, so you get water in the amount of 4*3.84 moles water, in round numbers, that is about 16*18 grams water, far more than you figured.
c. 4*3.84*18 *.687 = what
You have made some progress in solving the problem, but there are a few errors and misconceptions. Let's go through each part of the problem and correct any mistakes:
A. To determine the limiting reactant, we need to compare the moles of each reactant. You correctly calculated the number of moles of propane (C3H8).
However, your calculation for the number of moles of oxygen is incorrect. You stated that for 1 mole of C3H8, there are 5 moles of O2. This is inaccurate because the balanced chemical equation shows that 5 moles of O2 react with 1 mole of C3H8.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
Therefore, for every 1 mole of C3H8, we need 5 moles of O2. Let's recalculate:
150 g C3H8 x (1 mol C3H8 / 44.1 g C3H8) = 3.40 moles C3H8
3.40 moles C3H8 x (5 moles O2 / 1 mole C3H8) = 17.0 moles O2
Comparing the moles of C3H8 and O2, we see that 3.40 moles of C3H8 require 17.0 moles of O2. Since we have an excess of O2, the limiting reactant is C3H8.
B. To determine the number of grams of water produced, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 1 mole of C3H8 produces 4 moles of H2O. Let's calculate:
3.40 moles C3H8 x (4 moles H2O / 1 mole C3H8) = 13.6 moles H2O
Now, we need to convert moles of water to grams. The molar mass of water (H2O) is 18.0 g/mol.
13.6 moles H2O x (18.0 g H2O / 1 mole H2O) = 244.8 g H2O
Therefore, the number of grams of water produced is 244.8 g.
C. To calculate the number of grams of water obtained based on the given percent yield, we need to use the actual yield and the theoretical yield.
The actual yield is the amount of water obtained in the laboratory. The theoretical yield is the maximum amount of water that could be obtained based on stoichiometry.
You stated that the percent yield was 68.7%. Let's assume the actual yield is X grams.
Percent yield = (X / Theoretical yield) x 100
Theoretical yield is the amount of water calculated in part B: 244.8 g.
68.7 = (X / 244.8) x 100
Solving for X:
X = (68.7 / 100) x 244.8
X = 168.0 g
Therefore, the number of grams of water obtained in the laboratory, given a 68.7% percent yield, is 168.0 g.
In summary:
A. The limiting reactant is C3H8.
B. The number of grams of water produced is 244.8 g.
C. The number of grams of water obtained in the laboratory is 168.0 g (assuming a 68.7% percent yield).