Question
Asked Oct 3, 2019
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0.349 g of CaCl2•2H2O and 0.698 g of Na2CO3 are dissolved in 100 mL of water to form
a solution.
a. What is the limiting reactant?
b. How many grams of CaCO3 will precipitate?
c. How many grams of the excess reactant remain after the reaction has gone to
completion?

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Expert Answer

Step 1

(a) The resulting equation is given below.

CaCl.2H2O+Na,CO,CaCO, +2 NaCl +2H,O
From the above balanced chemical equation, it can be stated that one mole of
CaCl.2H,O is reaction with one mole of Na,CO, to give one mole CaCO;, two
mole of NaCl and two moles of water
But actual moles of CaCl,.2H2O can be calculated as given below
0.349g
mass
- 0.002734 mol.
Molecular mass
147.015g /mol
Similarly, for Na,cO, the actual moles can be calculated as given below.
0.698 g
mass
0.006586 mol.
Molecular mass
105.99g /mol
By comparing the moles of both the reactant it can be seen that Na2CO3 is in excess. So
the limiting reagent is CaCl.2H2O.
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CaCl.2H2O+Na,CO,CaCO, +2 NaCl +2H,O From the above balanced chemical equation, it can be stated that one mole of CaCl.2H,O is reaction with one mole of Na,CO, to give one mole CaCO;, two mole of NaCl and two moles of water But actual moles of CaCl,.2H2O can be calculated as given below 0.349g mass - 0.002734 mol. Molecular mass 147.015g /mol Similarly, for Na,cO, the actual moles can be calculated as given below. 0.698 g mass 0.006586 mol. Molecular mass 105.99g /mol By comparing the moles of both the reactant it can be seen that Na2CO3 is in excess. So the limiting reagent is CaCl.2H2O.

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Step 2

(b) 0.002734 mol of CaCl2.2H2O will produce 0.002734 mol of CaCO3 according...

So the mass of CaCO3 can be calculated as given below
Mass of CaCO3 =no.of molesx Molecular mass of CaCO
0.002374molx 100.0869 g / mol
0.238g
Hence, 0.238 g of CaCO3 will precipitate
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So the mass of CaCO3 can be calculated as given below Mass of CaCO3 =no.of molesx Molecular mass of CaCO 0.002374molx 100.0869 g / mol 0.238g Hence, 0.238 g of CaCO3 will precipitate

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