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/mole4.Calculate the experimentally determined ratio of moles of NaCl produced to moles of Na2CO3 reacted according to: mole-ratio = (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 21. 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?
