Question

Using the balanced equation: 2 NH4Cl + Ca(OH)2 ---> CaCl2 + 2 NH3 + 2 H20If 85 g of NH4Cl reacts with Ca(OH)2, how much ammonia can be formed?

Answer 1

Answer: 27.08 grams of ammonia will be formed .

Solutions -

We are given :

• 2 NH4Cl + Ca(OH)2 ---> CaCl2 + 2 NH3 + 2 H20

• Mass of NH4Cl = 85 g

,

• Molar mass NH4Cl = 53,491 g/mol

1. Calculate number of moles for NH4Cl

moles NH4Cl = massNH4Cl / Molar Mass NH4Cl

= 85g/53.491g/mol

=1.59 moles of NH4Cl

2. Determine how much ammonia is formed through the mole ratio from the balanced equation .

• We can see that ,2 moles NH4Cl, reacts with 1 mole Ca(OH)2 to form ,2 mole NH3,

,

• 2 moles NH4Cl : 2 mole NH3

so, 1.59 molesNH4Cl : x mole NH3 .....( do a cross multiply)

∴ x mole NH3 = (1.59* 2 )/2

= 1.59 moles

• This means that 85 g NH4Cl will produce 1.59 moles of NH3.

3. Calculate mass of NH3

given : moles = 1.59

Molecular mass NH3 = 17,031 g/mol

Mass NH3 = mole * Molecula mass

= 1.59mol* 17.031g/mol

= 27.08 grams

This means that 27.08g of Ammonia will be formed.