K2CO3 + 2HC1 -- H2O + CO2 + 2KC1 what would be the percent yield if you reacted 34.5g of K2CO3 and 22.5g of HC1 and produced 3.4 g of H2O?
K2CO3 + 2HC1 -- H2O + CO2 + 2KC1 what would be the percent yield if you reacted 34.5g of K2CO3 and 22.5g of HC1 and produced 3.4 g of H2O?
This is a limiting reagent (LR) problem as well as the percent yeild.
K2CO3 + 2HCl -- H2O + CO2 + 2KCl
mols K2CO3 = grams/molar mass = approx 0.25
mols H2O produced if K2CO3 is the LR = approx 0.25
mols HCl = g/molar mass = 0.6
mols H2O produced if HCl is the LR = 0.3
The LR is the smaller of the two which is K2CO3 or approx 0.25 mols H2O.
g H2O produced = mols x molar mass = approx 0.25 x 18 = about 4.5 g which is the theoretical yield (TY), The actual yield is 3.4 g (AY)
%yield = (AY/TY)*100 =100(3.4/4.5) = ?
All of those numbers are estimates and you should go through all of the calculations to confirm a better number. Post your work if you get stuck.
I'm looking for the answer
To calculate the percent yield, we need to compare the actual yield (the amount of H2O produced in the reaction) with the theoretical yield (the maximum amount of H2O that can be produced based on the given amounts of reactants).
In this balanced chemical equation:
K2CO3 + 2HCl → H2O + CO2 + 2KCl
We can see that the stoichiometric ratio between K2CO3 and H2O is 1:1. This means that the molar ratio is 1:1 as well, and the number of moles of K2CO3 is equal to the number of moles of H2O.
Step 1: Calculate the number of moles of K2CO3 and HCl:
Molar mass of K2CO3 (potassium carbonate) = 2 * (39.1 g/mol) + 12.0 g/mol + 3 * (16.0 g/mol) = 138.2 g/mol
Molar mass of HCl (hydrochloric acid) = 1 * (1.0 g/mol) + 35.5 g/mol = 36.5 g/mol
Number of moles of K2CO3 = mass of K2CO3 / molar mass of K2CO3 = 34.5 g / 138.2 g/mol
Number of moles of HCl = mass of HCl / molar mass of HCl = 22.5 g / 36.5 g/mol
Step 2: Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation. The reactant that produces the fewer moles of H2O will be the limiting reactant.
According to the balanced equation, the stoichiometric ratio between K2CO3 and H2O is 1:1. So, the moles of H2O that can be produced from K2CO3 = Number of moles of K2CO3.
The moles of H2O that can be produced from HCl = 2 * Number of moles of HCl (based on the stoichiometric ratio).
Step 3: Calculate the theoretical yield of H2O:
The theoretical yield is the amount of H2O that can be produced based on the limiting reactant.
Theoretical yield of H2O = moles of H2O produced from the limiting reactant * molar mass of H2O
Step 4: Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) * 100
Let's plug in the values to calculate the percent yield:
Number of moles of K2CO3 = 34.5 g / 138.2 g/mol = 0.25 mol
Number of moles of HCl = 22.5 g / 36.5 g/mol = 0.62 mol
Moles of H2O that can be produced from K2CO3 = 0.25 mol
Moles of H2O that can be produced from HCl = 2 * 0.62 mol = 1.24 mol
The limiting reactant is K2CO3 since it produces fewer moles of H2O.
Theoretical yield of H2O from K2CO3 = 0.25 mol * 18.0 g/mol = 4.5 g
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (3.4 g / 4.5 g) * 100 = 75.6%
Therefore, the percent yield would be 75.6%.