Balancing Chemical Reactions

The video tutorial explains the decomposition reaction of sodium chlorate into sodium chloride and oxygen gas. It guides viewers through the process of balancing the chemical equation by adjusting the number of atoms on each side. The tutorial emphasizes the importance of achieving an even number of oxygen atoms to facilitate balancing. The final steps involve ensuring the correct number of sodium and chlorine atoms, resulting in a balanced equation. The video concludes with a summary of the balanced equation.

What type of reaction is sodium chlorate undergoing in this equation?

• Double replacement reaction

• Synthesis reaction

• Decomposition reaction

• Single replacement reaction

What are the products formed when sodium chlorate decomposes?

• Sodium hydroxide and oxygen gas

• Sodium and chlorine gas

• Sodium chloride and oxygen gas

• Sodium oxide and chlorine gas

Why is it important to get the number of oxygen atoms to an even number when balancing this equation?

• To make the equation simpler

• To ensure the equation is balanced

• To match the number of chlorine atoms

• To match the number of sodium atoms

What is the first step taken to balance the oxygen atoms in the equation?

• Multiply the oxygen by three

• Add more chlorine atoms

• Multiply the oxygen by two

• Add more sodium atoms

After balancing the oxygen atoms, what is the next step in balancing the equation?

• Adjust the number of sodium atoms

• Adjust the number of nitrogen atoms

• Adjust the number of hydrogen atoms

• Adjust the number of chlorine atoms

How many sodium atoms are present on the product side after balancing?

• One

• Four

• Two

• Three

What coefficient is placed in front of NaCl to balance the equation?

• 3

• 1

• 4

• 2

What is the final balanced equation for the decomposition of sodium chlorate?

• NaClO3 → 2 NaCl + 3 O2

• 2 NaClO3 → NaCl + 2 O2

• NaClO3 → NaCl + O2

• 2 NaClO3 → 2 NaCl + 3 O2