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.