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.

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.
<span>(0.1875 moles)(98.004 g/mole) = 18.37575 g </span>
<span>In correct number of significant figures: 18.4 </span>
<span>Answer:
(16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m
a.)
(3.9704 m) x (1.86 °C/m) = 7.38 °C change
0.00°C - 7.38 °C = - 7.38 °C
b.)
(3.9704 m) x (0.512 °C/m) = 2.03 °C change
100.00°C + 2.03 °C = 102.03 °C</span>