2Al(NO2)3(s) + 6NH4Cl(s) -------------------> 2AlCl3 + 6N2 + 12H2O no of moles o

Question

2Al(NO2)3(s) + 6NH4Cl(s) -------------------> 2AlCl3 + 6N2 + 12H2O

no of moles of Al(NO2)3   = W/G.M.Wt

                                           = 61.5/165    = 0.372 moles

no of moles of NH4Cl          = W/G.M.Wt

                                            = 49.6/53.5   = 0.927 moles

2 moles of Al(NO2)3 react with 6 moles of NH4Cl

0.372 moles of Al(NO3)2 react with = 6*0.372/2    = 1.116 moles of NH4Cl

    NH4Cl is limiting reactant

6 moles of NH4Cl react with Al(NO2)3 to gives 6 moles of N2

0.927 moles of NH4Cl react with Al(NO2)3 to gives = 6*0.927/6   = 0.927 moles of N2

mass of N2 gass = no of moles * gram molar mass

                              = 0.927*28    = 25.956g of N2 >>>>.answer

Explanation / Answer

Solid Aluminum nitrite reacts with solid ammonium chloride to form aluminum chloride, nitrogen gas, and water. What is the maximum possible mass of gas produced when 61.5g of aluminum nitrite and 49.6g of ammonium chloride are allowed to react completely.? Please show al steps so I can understand it. Thank you.

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