Balancing Chemical Equations for Gold(III) Oxide

The video tutorial explains how to balance the chemical equation for the decomposition of gold(III) oxide into gold and oxygen. It begins with an introduction to the concept of balancing equations, followed by counting the atoms on both sides of the equation. The tutorial then demonstrates the steps to balance the equation by adjusting coefficients, emphasizing the importance of achieving an even number of atoms. Finally, the video concludes with the finalized balanced equation and a summary of the process.

What is the initial chemical equation for the decomposition of gold(III) oxide?

• Au2O3 → Au2 + O

• AuO → Au + O2

• Au3O2 → Au + O2

• Au2O3 → Au + O2

How many gold atoms are present on the reactant side of the equation initially?

• One

• Two

• Three

• Four

What is the strategy used to balance the oxygen atoms in the equation?

• Add more gold atoms

• Add more oxygen molecules

• Remove oxygen atoms

• Multiply the entire equation by two

After balancing the oxygen atoms, how many oxygen molecules are on the product side?

• Three

• Two

• Four

• Six

What is the final step to balance the gold atoms in the equation?

• Add a coefficient of four to the gold on the product side

• Add a coefficient of two to the gold on the reactant side

• Remove one gold atom from the reactant side

• Add more oxygen molecules

What is the balanced chemical equation for the decomposition of gold(III) oxide?

• 2 Au2O3 → 2 Au + 2 O2

• 4 Au2O3 → 8 Au + 6 O2

• Au2O3 → 2 Au + 3 O2

• 2 Au2O3 → 4 Au + 3 O2

Why is it often easier to balance equations by converting odd numbers to even?

• It makes the equation more complex

• It ensures all coefficients are whole numbers

• It simplifies the calculation

• It reduces the number of atoms