1029g calcium nitrate and 1072g ammonium fluoride react comp

10.29g calcium nitrate and 10.72g ammonium fluoride react completely to form calcium fluoride, dinitrogen monoxide, and water vapor. What is the mass present of each of the 5 substances in the reaction?

Solution

Balanced equation for given process is:

Ca(NO3)2 + 2 NH4F CaF2 + 2 N2O + 4 H2O

Molar mass of Ca(NO3)2=164.09 gram/mol

Molar mass of NH4F=37.0369 gram/mol

Molar mass of CaF2=78.075 gram/mol

Molar mass of N2O=44.0128 gram/mol

Molar mass of H2O=18.0153 gram/mol

number of moles of Ca(NO3)2) in 10.29g= 10.29/164.09=0.0627

number of moles of NH4F in 10.72g= 10.72/37.0369=0.28944

0.0627 mole of Ca(NO3)2 would react completely with 0.0627 x (2/1) = 0.1254 mole of NH4F, but there is more NH4F present than that, so NH4F is in excess and Ca(NO3)2 is the limiting reactant.

Amount of NH4F left =((0.28944 mol NH4F intiallly) - (0.1254 mol NH4F reacted)) x(37.0369 g NH4F/mol) =6.076 gram

Amount of CaF2 =(0.0627mol Ca(NO3)2) x (1 mol CaF2 / 1 mol Ca(NO3)2) x (78.075g CaF2/mol) =4.8953 gram

Amount of N2O=(0.0627 mol Ca(NO3)2) x (2 mol N2O / 1 mol Ca(NO3)2) x (44.0128 g N2O/mol) =5.52 gram

Amount of H2O=(0.0627 mol Ca(NO3)2) x (4 mol H2O / 1 mol Ca(NO3)2) x (18.0153 g H2O/mol) =4.518 gram