What is the mole ration between BaCO3 and H2O?
If the students wants to perform the reaction to obtain barium acetate and she has 20 grams of barium carbonate and 2 mol of CH3CO2H, which of these reactants is the limiting reactant? Show your procedure to get to the answer
How many grams of barium acetate will the student get if we have a 100 % yield (theoretical yield) for this reaction?
she gets 0.05 g of barium acetate instead, what would the % yield for this reaction?
How many mol of CO2 and H2O will she get from this reaction assuming a 100% yield?
BaCO3 + 2 CH3COOH = Ba(CH3COO)2 + H2O + CO2
1 mole of BaCO3 reacting with 2 moles of CH3COOH to produce 1 mol of barium acetate
mole of BaCO3 = 20 / 197.34 = 0.10135 mole
sotichiometic moles of BaCO3 = 0.10135 moles
mole of CH3COOH = 2 mol
stoichiometric moles of CH3COOH = 2/ 2 = 1 mole
since the number of moles of BaCO3 are lesser than the moles of CH3COOH so the BaCO3 is limiting reagent.
for 100% yield
moles of barium acetate produced = 0.10135 moles
gram of barium acetate produced = 0.10135* 255.43 = 25.9 g
percentage yield = (0.05 / 25.9)*100 = 0.19 %
moles of CO2 and H2O for 100% yield
1 mole of BaCO3 producing 1 mole of CO2 and 1 mole of H2O
so moles of CO2 produced = 0.10135 moles
moles of H2O produced = 0.10135 moles
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