Question

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. What mass of ammonium flouride is present after 26.24g calcium nitrate and 27.34g ammonium fluoride react completely?

Answer 1

Balanced equaiton is,

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

Molar mass of calcium nitrate = 164 g/mol

Molar mass of ammonium fluoride = 37 g/mol

From the balalnced equation,

1 mol calcium nitrate = 2 mol ammonium fluoride

164 g. of calcium nitrate needs 2 * 37 = 74 g. of ammonium fluoride

then, 26.24 g. of calclium nitrate needs 26.24 * 74 / 164 = 11.84 g. of ammonium fluoride.

SO, remaining ammonium fluoride after the reaction = 27.34 - 11.84 = 15.50 g.

But, we have 0.320 mol ammonium fluoride. Hence it is excess reagent and calciumnitrate is limiting reagent.

Remaing moles of