Asked by Glenda

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?
Answered by Glenda
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?
Answered by DrBob222
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.
Answered by Austin Den Dooven
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