Question

Ammonium chloride and calcium hydroxide react according to the following balanced equation:

2 NH₄Cl_(aq)  + Ca(OH)_2(aq)  -->  CaCl_2(aq)  + 2 NH_3(g)  + 2 H₂O_(l)

How many grams of CaCl₂ will be produced by completely reacting 1.4 L of a 2.33 M NH₄Cl solution react with excess Ca(OH)₂?

Answer 1

Moles of NH₄Cl = $1.4×2.33= 3.262 mole$

Moles CaCl $=\frac{3.262}{2}=1.631 moles$

Mass = $1.631×111= 181.04 g$