Question

stoichiometry and a precipitation reaction. Need to fill in the data table. The following information is given. 1.5 gram of CaCL2*2H2O

How many moles are present of CaCL2*2H2O and then calculate how many moles of pure CaCL2 are present in 1.5g of CaCL2*2H2O

Initial CaCL2*2H2O(g) = ???

Initial CaCL2*2H2O(moles) =??

Initial CaCL2(moles) = ???

Use the information given - how many moles of Na2CO3 are necessary to reach stoichiometry Quantities. From the calculation determine how many grams of NA2CO3 are necessary to reach stoichiometry quantities.

NaCO3 (moles) =????

Na2CO3 (g)=????

Theoretical CaCO3(g)=???

Answer 1

given 1.5g of CaCL2*2H2O

Initial CaCL2*2H2O(g) = 1.5 g

Initial CaCL2*2H2O(moles) = weight / molarmass

      weight = 1.5 g

molarmass of CaCL2*2H2O = 147 g/mol

= 1.5/147 = 0.0102 mol

1 mol CaCl2*2H2O = 1 mol CaCl2

Initial CaCL2(moles) = 0.0102 mol

equation: CaCl2 + Na2CO3 -----> CaCO3(S) + 2 NaCl

from equation: 1 mol CaCl2 = 1 mol Na2CO3 = 1 mol CaCO3

no of mol of CaCl2 reacted = 0.0102 mol

no of mol of NaCO3 (moles) required = 0.0102 mol

no of mol of Na2CO3 (g) required = n*Molarmass

                                                          = 0.0102*106

                                                         = 1.0812 g

no of mol of CaCO3 (g) formed = 0.0102 mol

Theoretical yield of CaCO3(g)= 0.0102*100 = 1.02 g