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
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!
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