2NH4CL + Ca(OH)2 = 2NH3 + 2H2O + CaCL2 nNH4CL = 3/53,5 = 0,056mol nCa(OH)2 = 3,7/74 = 0,05mol nNH3 = nNH4CL = 0,56mol mNH3 = 0,056 x 17 = 0,952g
2NH4CL + Ca(OH)2 = 2NH3 + 2H2O + CaCL2
nNH4CL = 3/53,5 = 0,056mol
nCa(OH)2 = 3,7/74 = 0,05mol
nNH3 = nNH4CL = 0,56mol
mNH3 = 0,056 x 17 = 0,952g