Sample Questions:

1. (JUNE 2000 416): Which of the following chemical equations is balanced correctly?
1. CH₃COOH + 2 NaOH --> CH₃COONa + 2 H₂O
2. CH₃COOH + NaOH --> 2 CH₃COONa + H₂O
3. CH₃COOH + NaOH --> CH₃COONa + H₂O
4. 2CH₃COOH + NaOH --> CH₃COONa + 2 H₂O

 

Answer C

2. (JUNE 2000 416): One of the causes of acid rain is the sulfur released when fossil fuels such as coal and oil are burned. The following two reactions take place when these fuels are burned:

sulfur dioxide + oxygen --> sulfur trioxide

sulfur trioxide + water --> sulfuric acid

With these reactions in mind, a student combined 128 g of sulfur dioxide with 32 g of oxygen to produce sulfur trioxide. He then combined all the resulting sulfur trioxide with 36 g of water to make sulfuric acid.

What mass of sulfuric acid did he produce?

Answer (128 + 32) + 36 = 196 g

3. (JUNE 1999 416): A reaction involving 168 g of sodium bicarbonate and 120 g of vinegar produces 88 g of carbon dioxide, 36 g of water and some salt.

How much salt is produced?

1. 80 g
2. 164 g
3. 340 g
4. 412 g

Answer B

4. (JUNE 1998 416): Balance the following equation correctly: Fe₂O₃ + C --> CO₂ + Fe

Notes:

When balancing equations remember these simple rules:

1. Never touch the small numbers (the subscripts).
2. Only introduce coefficients (large numbers) to create, for each element, the same total number of atoms on each side of the arrow.
3. Each coefficient applies to every atom in the compound. Ex. 4 HNO₃ means there are 4 H's, 4 N's and 4(3) =12 O's.
4. For mass problems, the total mass of the reactants (left hand side) equals the total mass of the products.

Using example 4 from June 1998, Fe₂O₃ + C --> CO₂ + Fe:

The solution was 2 Fe₂O₃ + 3 C --> 3CO₂ + 4Fe

Notice that there were three O's on the left-hand side but only two on the right. A common multiple of 2 and 3 is 6. As a result we created 6 on each side by using coefficients of 2 and 3. But by doing so, we created 4 Fe's. So we had to fix the Fe on the right hand side by using a 4. Finally we balanced the carbons by placing a "3" in front of the carbon.

Useful trick: If there is an element in the equation, leave that to the end, because any coefficient you introduce in front of that element will only affect that element.