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calcium nitrate and ammonium fluoride e http//eztio mheducation.com/hm.tpx ter 3

ID: 567192 • Letter: C

Question

calcium nitrate and ammonium fluoride e http//eztio mheducation.com/hm.tpx ter 3- Stoichiometry XPC Laptops connect. CHEMITIA CRN20407 SPRING20 20407 HEMISTRY hapter 3-Stoichiometry Question 36 (of 39) 36. 100 points 3 out of 12 attempts Be sure to answer all parts. Assistance Calcium nitrate and ammonium fuoride react to form calcium fluoride, dinitrogen mosoxide, and water vapor. What mass of each sabstance is present after 17.20gof calcium nitrate and 17.92 gof ammonium fluoride react completely? 4 1 g calcium nitrate | | g ammonium nuoride dinitrogen monoride g calcium fuoride water View Details of Last Check Answer Type here to seatdh

Explanation / Answer

Calcium nitrate and ammonium flouride will react according to following reaction

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

Moles Ca(NO3)2 = 17.2 g / 164 g/mol = 0.105

Moles NH4F = 17.92 g / 37.04 g/mol = 0.484

according to reaction stoichiometry 1 mole of calcium nitrate will react with 2 moles of ammonium fluoride

for 0.105 moles of calcium nitrate will require 0.105 * 2 = 0.210 moles of ammonium fluoride

Calcium nitrate is the limiting reactant

Moles NH4F in excess = 0.484 - 0.210 = 0.274

Mass NH4F in excess = 0.274 mol * 37.04 g/mol = 10.15 g Answer

Moles CaF2 produced = 0.105 mol

Molar mass of CaF2 = 78.1 g/mol

Mass of CaF2 = 0.105 mol * 78.1 g/mol = 8.2 g Answer

Moles N2O produced = 0.105 * 2 = 0.210 moles

Molar mass of N2O = 44 g/mol

mass of N2O = 0.210 moles * 44 g/mol = 9.24 g Answer

Moles water produced = 0.105 * 4 = 0.420 mol

Molar mass of water = 18 g/mol

Mass of water = 0.420 moles * 18 g/mol = 7.56 g Answer

Mass of ammonium fluoride = 0 g Answer

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