Question

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. What mass of each substance is present after 10.61 g of calcium nitrate and 11.06g of ammonium fluoride react completely?

Answer 1

Ca(NO3)2 (aq) + 2NH4F-----------> CaF2 + 2N2O + 4H2O

no of moles of Ca(NO3)2   = W/G.M.Wt   = 10.61/164   = 0.065moles

no of moles of NH4F        = W/G.M.Wt     = 11.06/37    = 0.298 moles

2 moles of NH4F react with 1 moles of Ca(NO3)2

0.298 moles of NH4F react with = 1*0.298/2   = 0.149 moles of Ca(NO3)2 is required

Ca(NO3)2 is limiting reactant

Ca(NO3)2 (aq) + 2NH4F-----------> CaF2     +          2N2O        +    4H2O

0.065 moles                                    0.065 moles      2*0.065 moles      4*0.065 moles

mass of CaF2 = no of moles * gram molar mass

                        = 0.065*78   = 5.07g

mass of N2O   = no of moles * gram molar mass

                       = 2*0.065*44   = 5.72g

mass of H2O = no of moles * gram molar mass

                       = 4*0.065*18   = 4.68g