Audrey carried out the following reaction in lab: CaCl_2(aq) + Na_2CO_3(aq) righ
ID: 492984 • Letter: A
Question
Audrey carried out the following reaction in lab: CaCl_2(aq) + Na_2CO_3(aq) rightarrow CaCo3_(s) + 2NaCl_(aq) Audrey weighs out 1.115 g of Na2CO3 and 0.719 g of CaCI2, and dissolves both substances in water before mixing the two solutions together. What is the maximum amount of precipitate that she should hope to collect? What is the limiting reactant in the reaction described above? Occasionally, some students determine their percent yield to be greater than 100%. What experimental error could have led to this result, and what should be watched for to insure that you do not obtain this result when you perform the experiment? After performing Trial 1, Audrey forgot to empty the liquid from the filtration flask before filtering the product from Trial 2. As soon as the liquid from Trial 2 entered the flask, a white precipitate formed. Why did this happen?Explanation / Answer
CaCl2 + Na2CO3 -------->CaCo3 + 2NaCl
no of moles of Na2CO3 = W/G.M.Wt = 1.115/106 = 0.0105 moles
no of moles of CaCl2 = W/G.M.Wt = 0.719/111 = 0.0065 moles
from balanced equation
1 moles of CaCl2 react with 1 mole of Na2Co3
0.0065 moles of CaCl2 react with 0.0065 moles of Na2Co3
CaCl2 is limiting reagent
1 mole of CaCl2 react with Na2Co3 to gives 1 mole of CaCO3
0.0065 moles of CaCl2 react with Na2CO3 to gives 0.0065 moles of CaCO3
mass of CaCO3 = no of moles * gram molar mass
= 0.0065*100 = 0.65 g
0.65g of CaCO3 precipitare is collected.
percent yield = Actual yield*100/theoritical yield
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