Question

Using the mass of baking soda that you used this lab, calculate the theoretical (expected) yield of NaCH3COO in grams ?

Answer 1

Answer:

4.4 g

Explanation:

The equation of the reaction is;

NaHCO3 (aq) + CH3COOH (aq) ----> CO2 (g) + H2O (l) + CH3COONa (aq)

Mass of baking soda - 4.5 g

Molar mass of NaHCO3 = 84.007 g/mol

Number of moles of baking soda= 4.5 g/84.007 g/mol = 0.0536 moles

If 1 mole of NaHCO3 yields 1 mole of CH3COONa

0.0536 moles of NaHCO3 also yields 0.0536 moles moles of CH3COONa

Hence;

Theoretical yield of CH3COONa =  0.0536 moles * 82.0343 g/mol = 4.4 g