Question

calcium nitrate and ammonium fluride react to form calcium fluoride, dinitrogen monoxide, and water vapor. what mass of each substance is present after 17.60 g of calcium nitrate and 18.34 g of ammonium fluoride react completely? (A) g of calcium nitrate (b) g of ammounium fluoride (c) calcium fluoride (D) dinitrogen monoxide (E) water

Answer 1

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

Molar mass of Ca(NO3)2 = 40 + (14+48)2 = 40 + 124 = 164 gm/mol

number of moles of Ca(NO3)2 = mass/molar mass = 17.60/164 = 0.1073 moles

Molar mass of NH4F = 14 + 4 * 1 + 19 = 37 gm/mol

Number of moles of NH4F = 18.34/37 = 0.49567 moles

Hence Ca(NO3)2 is the limiting reagent

moles of CaF2 = 0.1073 moles

moles of N2O = 2 * 0.1073 moles

moles of H2O = 4 * 0.1073 moles

moles of Ca(NO3)2 = 0

moles of NH4F = 0.49567 - 2 * 0.1073 moles

Number of moles of ammonium flouride