1) 2.00 grams of H₂O

Step One: the problem gives 2.00 grams

Step Two: the molar mass of H₂O is 18.015 grams/mole

Step Three: 2.00 grams ÷ 18.015 grams/mole = 0.111 mole (when rounded off to the correct number of significant figures)

The three steps, put together, look like this:

Set up as a proportion, the problem looks like this:

2) 75.57 grams of KBr

Step One: the problem gives 75.57 grams

Step Two: the molar mass of KBr is 119.0023 grams/mole

Step Three: 75.57 grams ÷ 119.0023 grams/mole = 0.6350 mole (when rounded off to the correct number of significant figures)

The three steps, put together, look like this:

Set up as a proportion, the problem looks like this:

3) 100. grams of KClO₄

Step One: the problem gives 100. grams (note three significant figures)

Step Two: the molar mass of KClO₄ is 138.549 grams/mole

Step Three: 100. grams ÷ 138.549 grams/mole = 0.722 mole (when rounded off to the correct number of significant figures)

The three steps, put together, look like this:

Set up as a proportion, the problem looks like this:

4) 8.76 grams of NaOH

Step One: the problem gives 8.76 grams

Step Two: the molar mass of NaOH is 39.997 grams/mole

Step Three: 8.76 grams ÷ 39.997 grams/mole = 0.219 mole (when rounded off to the correct number of significant figures)

The three steps, put together, look like this:

Set up as a proportion, the problem looks like this:

5) 0.750 grams of Na₂CO₃

Step One: the problem gives 0.750 grams (note three significant figures)

Step Two: the molar mass of Na₂CO₃ is 105.989 grams/mole

Step Three: 0.750 grams ÷ 105.989 grams/mole = 7.08 x 10¯³ mole (when rounded off to the correct number of significant figures)

Please note that 7.08 x 10¯³ can be expressed as 0.00708.

The three steps, put together, look like this:

Set up as a proportion, the problem looks like this:

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