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Liquid octane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yield of water formed from the reaction of 6.85 g of octane and 46.3 g of oxygen gas?rnrnround your answer to 3 significant figures.
Accepted Answer
First, we need to balance the chemical equation:
C8H18 + 12.5 O2 → 8 CO2 + 9 H2O
To determine the limiting reactant, we will convert the given masses of octane and oxygen to moles using their respective molar masses:
moles of octane = (6.85 g) / (114.23 g/mol) = 0.06 mol
moles of oxygen = (46.3 g) / (32.00 g/mol) = 1.45 mol
Next, we will calculate the mole ratio between octane and oxygen based on the balanced equation:
mole ratio of octane to oxygen = 1 mol C8H18 / 12.5 mol O2
Comparing the mole ratio with the calculated moles of each reactant, we find that oxygen is in excess (1.45 mol > 0.06 mol * 12.5). Therefore, octane is the limiting reactant.
Using the mole ratio and the limiting reactant, we can calculate the theoretical yield of water:
moles of water produced = (0.06 mol C8H18) * (9 mol H2O / 1 mol C8H18) = 0.54 mol
Finally, we convert moles of water to grams using its molar mass:
theoretical yield of water = (0.54 mol) * (18.015 g/mol) = 9.73 g
Rounding to 3 significant figures, the theoretical yield of water formed from the reaction is 9.73 g.
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