Question
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 43.9 g of hexane is mixed with 57. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to significant digits.
Answers
This is a limiting reagent (LR) problem and I do these the long way.
Write and balance the equation.
2C6H14 + 19O2 ==>12CO2 + 14H2O
Convert 57 g O2 to mols. mol = grams/molar mass = ?
Do the same to convert 43.9 g hexane to mols.
Using the coefficients in the balanced equation, convert mols hexane to mols CO2
Do the same to convert mols O2 to mols CO2.
It is likely that these two values will not agree which means one is wrong. The correct value in LR problem is ALWAYS the smaller value and the reagent responsible for producing that value is the LR.
After you know the LR, you can determine how much of the excess reagent (the one that isn't the LR). Again, use the coefficients in the balanced equation to convert mols of the LR to mols of the excess reagent. Then to find how much is left, it is
initial amount - amount needed to react with the LR = amount in excess.
Post your work if you get stuck.
Write and balance the equation.
2C6H14 + 19O2 ==>12CO2 + 14H2O
Convert 57 g O2 to mols. mol = grams/molar mass = ?
Do the same to convert 43.9 g hexane to mols.
Using the coefficients in the balanced equation, convert mols hexane to mols CO2
Do the same to convert mols O2 to mols CO2.
It is likely that these two values will not agree which means one is wrong. The correct value in LR problem is ALWAYS the smaller value and the reagent responsible for producing that value is the LR.
After you know the LR, you can determine how much of the excess reagent (the one that isn't the LR). Again, use the coefficients in the balanced equation to convert mols of the LR to mols of the excess reagent. Then to find how much is left, it is
initial amount - amount needed to react with the LR = amount in excess.
Post your work if you get stuck.
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