Question

Ammonia gas can be produced by heating together the solid NH4Cl and Ca (OH) 2

Answer 1

Explanation:

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2O

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.107 g NH4Cl produces 34 g NH3.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.107 g NH4Cl produces 34 g NH3.100 g NH4Cl produces (34/107)×100 = 31.776 g NH3.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.107 g NH4Cl produces 34 g NH3.100 g NH4Cl produces (34/107)×100 = 31.776 g NH3.At STP, 17 g NH3 occupies 22.4L.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.107 g NH4Cl produces 34 g NH3.100 g NH4Cl produces (34/107)×100 = 31.776 g NH3.At STP, 17 g NH3 occupies 22.4L.31.776 g NH3 occupies (22.4/17)×31.776 = 41.87 L.

2 NH4Cl + Ca(OH)2 →2 NH3 + CaCl2 +2 H2OMolar mass of NH4Cl = 14+4×1+35.5 = 53.5 g/mol.Molar mass of Ca(OH)2 = 40+2×(16+1) =74 g/mol.From the stoichiometric equation we get, mass ratio, NH4Cl : Ca(OH)2 = (2×53.5) : 74 = 107 : 74 = 1.446 : 1.Given mass ratio, NH4Cl : Ca(OH)2 = 100 : 100 = 1 : 1.As the given mass of NH4Cl is in less amount, it is the limiting reagent here.107 g NH4Cl produces 34 g NH3.100 g NH4Cl produces (34/107)×100 = 31.776 g NH3.At STP, 17 g NH3 occupies 22.4L.31.776 g NH3 occupies (22.4/17)×31.776 = 41.87 L.Hope, this helps.

Answer 2

Answer:

Word equation:

State which characteristic is observed in the above reaction. Give an example of a similar above characteristic, see. between two other gaseous reactants