Calcium Hydroxide reacts with ammonium chloride to give ammonia according to the equation

Calcium Hydroxide reacts with ammonium chloride to give ammonia
according to the equation.
Ca(OH)₂+2 NH₄Cl → CaCl₂+2NH₃+2H₂0
If 5.35 g of ammonium chloride is used calculate :
(a) the weight of calcium chloride formed.
(b) the volume at STP of ammonia liberated at the same time.
[[H =1, N = 14, 0 =16, Cl = 35.5, Ca =40].

Report

Question

Given, 5.35 g of Ammonium chloride NH₄Cl reacts with Ca(OH)₂to form one mole of CaCl₂

Here, Molar mass of NH₄Cl = 5.35 g/mole

Molar mass of CaCl₂ = 111 g/mole

Molar mass of NH₃ = 17 g/mole

Therefore, no. of moles of NH₄Cl = Given massMolecular mass

= 5.3553.5

= 0.1 moles

Now, the no. of moles of CaCl₂ = 0.1 12

= 0.05 moles

Therefore, the mass of CaCl₂ will be = 0.05111

= 5.55 g

0.1 moles of NH₄Cl produce = 0.12

= 0.05 moles of NH₃.

Again, one mole of Ammonia (NH₃) at STP = 22.4 L

Therefore, 0.05 moles of Ammonia at STP = 0.0522.4

= 1.12 L

Hence, the answer of (a) is 5.55 g

(b) is 1,12 L

solved 5

wordpress 4 mins ago 5 Answer 70 views +22