2. Convert the mass of Na₂CO₃ to moles, given its molecular weight (MW) of 105.989 g/mole.

3. Convert the mass of NaCl to moles, given its molecular weight (MW) of 58.443 g/mole

4.Calculate the experimentally determined  ratio of moles of NaCl produced to moles of Na₂CO₃ reacted, according to:  mole-atio = (moles NaCl)/(moles Na₂CO₃)

5. Use the mole ratio,  from question#4,  to write a balanced chemical equation for the reaction.

6. What is the theoretical mole ratio according to this balanced equation?

7. Use the theoretical mole ratio to calculate the theoretical yield of NaCl in grams from 2 g of Na₂CO₃?

8. The "percent yield" is the ratio of the actual amount of a product to the theoretical amount. Calculate the percent yield of NaCl as:

% yield = (experimental grams NaCl) / (theoretical grams NaCl) * 100%

PART 2

1. Based on the balanced chemical equation, given that 2.0 g of Na₂CO₃(molecular weight = 58.44 g/mol) was added to the reaction, how many mL of H₂O (molecular weight = 18.00 g/mol) were produced?  Remember that moles = mass/molecular weight and that 1 g of H2O = 1 mL.2. What was the acid in the reaction?

3. What was the base in the reaction?

4. What was the salt formed in the acid-base reaction?

5. How does Na2CO₃ act as a base?