A 2.556g sample of baking soda (NaHCO3)decomposes to produce 0.876g of solid sodium carbonate. Caculate the theoretical yield and the percent yield of the sodium carbonate. The equation is:

Question

A 2.556g sample of baking soda (NaHCO3)decomposes to produce 0.876g of solid sodium carbonate. Caculate the theoretical yield and the percent yield of the sodium carbonate. The equation is:
2NaHCO3(s) -> Na2CO3(s) + H2O + CO2(g)

DrBob222 · Answer

This is a straight forward stoichiometry problem.
2NaHCO3 ==> Na2CO3 + CO2 + H2O
mols NaHCO3 = grams/molar mass = approx (estimated) 0.03
Convert to mols Na2CO3. The ratio in the equation is 2:1; therefore, mols Na2CO3 formed is estimated 0.015
grams Na2CO3 formed is g = mols x molar mass = estimated 1.6 g. This is the theoretical yield (TY)
Actual yield (AY) in the problem is 0.876g
% yield = (AY/TY)*100 = >
Check my work.

frank · Answer

thank you, i forgot to convert the mols, once i put that part in. I got about the same thing as you