Question

For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of carbon dioxide. benzene (C6H6) (l) oxygen (g) carbon dioxide (g) water (g) What is the theoretical yield of carbon dioxide

Answer 1

The theoretical yield of carbon dioxdide = 5.11 grams

Step-by-step explanation:

Step 1: Data given

Mass of oxygen gas = 4.64 grams

Molar mass of O2 = 32.0 g/mol

The reaction yields 3.95 grams of carbon dioxide (CO2)

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles oxygen

Moles oxygen = mass oxygen / molar mass oxygen

Moles oxygen = 4.64 grams / 32.0 g/mol

Moles oxygen = 0.145 moles

Step 4: Calculate moles of carbon dioxide (CO2)

For 2 moles C6H6 we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.145 moles O2 we'll have 12/15 * 0.145 = 0.116 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.116 moles * 44.01 g/mol

Mass CO2 = 5.11 grams

The theoretical yield of carbon dioxdide = 5.11 grams

Answer 2

Theoretical yield for CO₂ is 5.10g

Step-by-step explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g