Answer:
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
