Question

If 2.28 g C8H18 reacts with 7.00 g of Oz, predict the mass of CO, that can be produced and

identify the limiting reactant. Explain how you got your answers.
Unbalanced equation: C8H180 + 026) • CO26) + H₂O

Answer 1

O₂ is limiting reactant and 6.17g CO₂ can be produced

Step-by-step explanation:

Based on the balanced reaction:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

1 mole of C₈H₁₈ reacts with 25/2 moles O₂.

To solve this question we must convert the mass of each reactant to moles and, using the balanced reaction, we can find limiting reactant. With moles of limiting reactant we can find the moles of CO₂ and its mass as follows:

Moles C₈H₁₈ -Molar mass: 114.23g/mol-

2.28g * (1mol / 114.23g) = 0.0200 moles C₈H₁₈

Moles O₂ -Molar mass: 32g/mol-

7.00g * (1mol / 32g) = 0.219 moles O₂

For a complete reaction of 0.0200 moles C₈H₁₈ are needed:

0.0200 moles C₈H₁₈ * (25/2moles O₂ / 1mole C₈H₁₈) = 0.250 moles of O₂ are needed.

As there are just 0.219 moles, O₂ is limiting reactant.

Moles of CO₂ that can be produced:

0.219 moles O₂ * (8moles CO₂ / 25/2moles O₂) = 0.140 moles CO₂

The mass is -Molar mass CO₂: 44.01g/mol-:

0.140 moles CO₂ * (44.01g / mol) =

6.17g CO₂ can be produced