Question

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

Answer 1

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