Answer:
The product made is Na2CO3.
The % yield is 91.8 %
Explanation:
Step 1: Data given
Mass of baking soda (NaHCO3) = 5.0 grams
Molar mass of NaHCO3 = 84.0 g/mol
Each of the equations calculated is left with 2.4g of NaOH, 1.86 g Na2O, and 3.18g Na2Co3.
Step 2: The balanced equations
2NaHCO3 → Na2CO3 + CO2 + H20
NaHCO3 → NaOH + CO2
2NaHCO3 → Na2O + 2CO2 + H20
Step 3: Calculate moles NaHCO3
Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3
Moles NaHCO3 = 5.0 grams / 84.0 g/mol
Moles NaHCO3 = 0.060 moles
Step 4: Calculate moles of products
For 2 moles NaHCO3 we'll have 1 mol Na2CO3
For 0.060 moles NaHCO3 we'lll have 0.060 / 2 = 0.030 moles Na2CO3
For 1 mol NaHCO3 we'll have 1 mol NaOH
For 0.060 moles NaHCO3 we'll have 0.060 moles NaOH
For 2 moles NaHCO3 we'll have 1 mol Na2O
For 0.060 moles NaHCO3 we'll have 0.030 moles Na2O
Step 5: Calculate mass of products
Mass = moles * molar mass
Mass of Na2CO3 = 0.030 moles * 105.99 g/mol = 3.18 grams
Mass of NaOH = 0.060 moles * 40.0 g/mol = 2.4 grams
Mass of Na2O = 0.030 moles *61.98 g/mol = 1.86 grams
Step 6: Calculate the percent yield
% yield = actual yield / theoretical yield
% yield Na2CO3 = (2.92 grams / 3.18 grams) *100% = 91.8 %
% yield NaOH = (2.92 grams / 2.4 grams ) *100% = 121.6 %
% yield of Na2O = (2.92 grams / 1.86 grams ) * 100% = 157 %
The product made is Na2CO3, the other reactions have a % yield greater than 100 %
= 0.1000 mol/L × 
= 0.086 00 mol/L