"Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 94. g of octane is mixed with 200. g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits."

36.9 g is the mass of octane that is left over

Explanation:

We state the reaction by the information given.

Octane as a reactant → C₈H₁₈

Oxygen as the other reactant → O₂

We have both masses, so we can predict the limiting reactant

We convert the mass to moles:

94 g / 114g/mol = 0.824 moles of octane

200 g / 32 g/mol = 6.25 moles of O₂

The reaction is:

2C₈H₁₈(l) + 25O₂(g) →  16CO₂(g) + 18H₂O(g)

2 moles of octane need 25 moles of O₂ to react

Then, 0.824 moles of octane may react with (0.824 . 25) /2 =  10.3 moles O₂

We have 6.25 moles and we need 10.3, then the O₂ is the limiting reagent

Octane is the excess reactant:

25 moles of O₂ may react with 2 moles of octane

Therefore, 6.25 moles of O₂ will react with (6.25 . 2) / 25 = 0.500 moles

We have 0.824 moles of octane and we only need 0.500, therefore

(0.824-0.500) = 0.324 moles are left over by the reaction.

We convert the moles to mass → 0.324 mol . 114g / 1 mol = 36.9 g

There would remain 36.9 grams of octane

Explanation:

Step 1: Data given

Mass of octane = 94.0 grams

Molar mass of octane = 114.23 g/mol

Mass of oxygen = 200 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

2 C8H18 + 25 O2→ 16 CO2 + 18 H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles octane = 94.0 grams / 114.23 g/mol

Moles octane = 0.823 moles

Moles oxygen = 200 grams / 32.0 g/mol

Moles oxygen = 6.25 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles oxygen to produce 16 moles CO2 and 18 moles H2O

The limiting reactant is oxygen. It will completely be consumed (6.25 moles). Octane is in excess. There will react 6.25 / 12.5 = 0.5 moles

There will remain 0.823 - 0.500 = 0.323 moles

Step 5: Calculate mass of octane

Mass octane = Moles octane * molar mass octane

Mass octane = 0.323 moles * 114.23 g/mol

Mass octane = 36.9 grams

There would remain 36.9 grams of octane

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