A solution of calcium nitrate. Ca(NO_3)_2. reacts with a solution of ammonium fl

Question

A solution of calcium nitrate. Ca(NO_3)_2. reacts with a solution of ammonium fluoride. NH_4 F. to form solid calcium fluoride. CaF_2. and ammonium nitrate, NH_4 NO_3 (aq). When 45.00 ml of 4.8724 times 10^-1 M Ca(NO_3)_2 solution was added to 60.00 mL of 9.9981 times 10^-1 M NH_4 F solution, 1.524 g of CaF_2 was isolated. Calculate the number of moles of Ca(NO_3)_2 originally present A solution of calcium nitrate. Ca(NO_3)_2 reacts with a solution of ammonium fluoride. NH_4 F. to form solid calcium fluoride. CaF_2 and ammonium nitrate. NH_4 NO_3(aq). When 45.00 mL of 4.8724 times 10^-1 M Ca(NO_3)_2 solution was added to 60.00 ml of 9.9981 times 10^-1 M NH_4 F solution. 1.524 g of CaF_2 was isolated. Calculate the number of moles of NH_4 F originally present A solution of calcium nitrate. Ca(NO_3)_2. reacts with a solution of ammonium fluoride. NH_4 F, to form solid calcium fluoride. CaF_2, and ammonium nitrate. NH_4 NO_3(aq). When 45.00 mL of 4 8724 times 10^-1 M Ca(NO_3)_2 solution was added to 60.00 mL of 9.9981 times 10^-1 M NH_4 F solution, 1.524 g of CaF_2 was isolated. Calculate the theoretical yield, in grams, of CaF_2

Explanation / Answer

Reaction taking place is:

Ca(NO3)2 + 2NH4F ---> CaF2 + 2NH4NO3.

Moles of calcium nitrate solution taken = Molarity*Volume = 0.045*0.48724 = 0.0219258

Moles of ammonium flouride solution taken = Molarity*Volume = 0.060*0.99981 = 0.0599886

According to reaction stoichiometry,

1 mole calcium nitrate produces 1 mole of calcium flouride. Since calcium nitrate is in limiting amount, so moles of calcium fluoride formed = 0.0219258

Theoretical yield = Moles*MW = 0.0219258*78.07 = 1.7117 grams

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