Question

How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)

Answer 1

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

$200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}$

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200.  * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.