Ammonia gas can be produced by heating together the solid NH4Cl and ca

1.	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
Answer» 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:

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

Answer»

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: