Question

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l) + oxygen (g) carbon monoxide (g) + water (g) What is the maximum amount of carbon monoxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Step-by-step explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams