Ammonia gas can be produced by heating together the solid NH4Cl and calcium hydroxide. If a mixture containing 100g of each of these solids is heated, how many grams of NH3 are produced? also find the volume of NH3 gas at STP. 2NH4Cl + Ca(OH)^2___》2NH3 + CaCl2 + 2H2O
Answers
Answer:Ca(OH)2 +2 NH4Cl -> CaCl2 + 2 NH3 + 2H2O.
Thus 1 mole of Ca(OH)2 requires 2 moles of NH4Cl.
number of moles of in 100 g of each of reactants is
for Ca(OH)2 100 g / 74g mol^-1 = 1.35 moles
for NH4Cl. 100 g÷ 53.5 g mol^-1= 1.87 moles
In this case, NH4Cl is a limiting reagent
1.87 moles of it will react with 1.35/2= 0.675 moles of Ca(OH)2 producing 1.35 moles of NH3 i.e. 34 g × 1.35 =45.9 g of it.
Since 1 mole of any gas occupies 22.4 L
1.35 × 22.4 L corresponds to 30.24 L .
Explanation:
31.62 grams of NH₃ are produced.
The volume of NH₃ gas at STP is 41.664 L.
Explanation:
2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
From the above reaction, 2 moles of ammonium chloride and 1 mole of calcium hydroxide produces 2 moles of ammonia.
The number of moles of ammonium chloride is:
⇒ 100/53.5 = 1.86 moles
The number of moles of calcium hydroxide is:
⇒ 100/74 = 1.35 moles
1 mole of calcium hydroxide reacts with 2 moles of ammonium chloride:
From reaction, 1.35 moles of calcium hydroxide requires 1.35 × 2 = 2.7 moles of ammonium chloride which is not accessible.
Hence, Calcium hydroxide is not limiting reagent.
2 moles of ammonium chloride reacts with 1 mole of calcium hydroxide :
So 1.86 of ammonium chloride requires 1.86/2 = 0.93 moles of calcium hydroxide which is available.
Hence, ammonium chloride is limiting reagent.
Unreacted moles of calcium hydroxide = 1.35 - 0.93 = 0.42 moles
Now, 2 moles of ammonia are formed from 2 moles of ammonium chloride.
Therefore 1.86 moles would be formed from 1.86 moles of ammonium chloride.
Weight of ammonia produced = 1.86 × 17 = 31.62 grams
1 mole of ammonia measures 22.4 L at STP
1.86 mole would measure 1.86 × 22.4 = 41.664 L at STP.