If 3.5 moles Ca(OH)2 are added to 380ml 1.3M NH4Cl how many moles of Ca(OH)2 will be left over?
NH4Cl + Ca(OH)2 --> NH3 + H2O + CaCl2
Moles of NH"_4"Cl =Molarity*volume
moles=0.38*1.3=0.494
Here NH"_4" Cl is limiting reagent because it have less number of moles than calcium hydroxide.
1 mole calcium hydroxide react with 2 mole of NH"_4" Cl so,
remaining mole of calcium hydroxide =3.5-0.247=3.253 mole
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