Please help me understand how to find the limiting reactant, the theoretical yie
ID: 899466 • Letter: P
Question
Please help me understand how to find the limiting reactant, the theoretical yield of CO2(mol) and the theoretical yield of CO2(grams).
I just completed this lab and here are my values.
I know that the Ratio is 1:1 from the balanced equation however I dont know how to find what the limiting reactant is from each case and then how to find the Theoretical yield of CO2 in mols and in grams
If you need further information please let me know and thanks!
CH3COOH (added) mass CH3COO (grams) calculated mols of CH3COOH Mass NaHCO3 (g) Mols of NaHCO3 1 5ml 8.3 .0034 0.1 .0011 2 5ml 8.2 .0034 0.2 .0023 3 5ml 8.2 .0034 0.3 .0035 4 5ml 8.2 .0034 0.4 .0047 5 5ml 8.1 .0033 0.5 .0059 6 5ml 8.2 .0034 0.9 .0107Explanation / Answer
the chemical reaction is:
CH3COOH + NaHCO3 => CH3COONa + CO2 + H20
analysing the stochiometric coefficients, we see that one mole of CH3COOH willr eact with one mole of NaHCO3 to give one mole of CO2, i.e equal moles of both compounds have to be present in order to complete tl take place upo the point till the lesser compound is exhausted.
Thats the limiting reagent.
In the above case the limiting reagent is NaHCO3 in 1 and 2. after that CH3COOH is the limiting reagent.
THe theoretical yield of CO2 equals to the no. of moles of limiting reagent.
to find the mass:
molar mass of CO2= 44 g
mass=no. of moles * molar mass
if you have any further queries, let me know