A student starts with 4.961 g of unknown mixture of Na2CO3 and CaCl2*2H2O. They
ID: 1052364 • Letter: A
Question
A student starts with 4.961 g of unknown mixture of Na2CO3 and CaCl2*2H2O. They collect 1.769 g of precipitate. The identified Na2CO3 as their limiting reagent. How many grams of CaCl2*2H2O did the unknown mixture originally contain? CHEM 105 Fall 2016 Limiting Reactant Q4 A student starts with 4 961 g of unknown mature o NajCOs and CaC 2Hy0. They collect 1 709g of preciptate They identified Na CO as ther iming reagen Part A How many grams of CaCl 2H0 did the uniknown miture orginaly contan? My Answers Ghve Up My Answers Giee UpExplanation / Answer
Write out the equation taking place:
Na2CO3 + CaCl2.2H2O -------> CaCO3 (s) + 2 NaCl + 2 H2O
The total weight of (Na2CO3 + CaCl2.2H2O) mixture is 4.961 g. Let the mixture contain x g Na2CO3.
As per the balanced stoichiometric equation above, there is a 1:1 molar ration between Na2CO3 and CaCO3. CaCO3 is insoluble in water and gets precipitated out.
The student obtained 1.769 g precipitate. Therefore, the weight of CaCO3 precipitated out = 1.769 g.
Molar mass of CaCO3 = 100.1 g/mol.
Therefore, moles of CaCO3 precipitated out = (1.769 g)*(1 mole/100.1 g) = 0.01767 mole.
Moles of Na2CO3 reacted = (0.01767 mole CaCO3)*(1 mole Na2CO3/1 mole CaCO3) = 0.01767 mole.
Molar mass of Na2CO3 = 105.9888 g/mol.
Therefore, mass of Na2CO3 reacted = (0.01767 mole)*(10.5.9888 g/1 mole) = 1.8728 g.
The mixture contains 1.8728 g 1.873 g Na2CO3.
Therefore, mass of CaCl2.2H2O in the mixture = (4.961 – 1.873) g = 3.088 g
Ans: The mixture contains 1.873 g Na2CO3 and 3.088 g CaCl2.2H2O.
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