Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. What mass of each substance is present after 21.75 g of calcium nitrate and 22.66 g of ammonium fluoride react completely

The answer to your question is  

  Ca(NO₃)₂  = 0 g

NH₄F           8.785 g

CaF₂            14.63 g

N₂O              16.5 g

H₂O              13.5 g

Explanation:

Data

mass of Ca(NO₃)₂ = 21.75g

mass of NH₄F = 22.66 g

Balanced chemical reaction

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

molar mass of Ca(NO₃)₂ = 40 + (14 x 2) + (16 x 6) = 116 g

molar mass of NH₄F = 14 + 4 + 19 = 2 x 37 = 74 g

Calculate the limiting reactant

theoretical Ca(NO₃)₂/ NH₄F = 116/74 = 1.57

experimental Ca(NO₃)₂/ NH₄F = 21.75/22.66 = 0.95

The limiting reactant is Ca(NO₃)₂

Excess reactant = NH₄F      

              116 g of Ca(NO₃)₂     74 g of  NH₄F

              21.75 g                     x

               x = (21.75 x 74)/116

               x = 13.875

mass of  NH₄F left = 22.66 - 13.875 = 8.785

mass of CaF₂

               116 g of Ca(NO₃)₂  78 g of CaF₂

                21.75 g                  x

                x = (21.75 x 78)/116

                x = 14.625 g

mass of N₂O

                116 g of Ca(NO₃)₂  88 g of N₂O

                21.75 g                    x

                x = (21.75 x 88)/116

               x = 16.5 g

mass of H₂O

                   116 g of Ca(NO₃)₂  72 g of water

                   21.75 g                    x

                   x = (21.75 x 72)/116

                  x = 13.5 g  

i think the answer is terrae but not 100 percent sure. ▓▬▬ §