Ammonia gas can be prepared by the reaction CaO(s) + 2NH4Cl(s) -> 2NH3(g) + H2O(g) + CaCl2(s) If 112 g CaO reacts with 224 g NH4Cl, how many moles of reactants and products are there when the reaction is complete?
To determine the number of moles of reactants and products in the given reaction, we need to use the molar masses of each substance.
The molar mass of CaO (calcium oxide) is 56.08 g/mol.
The molar mass of NH4Cl (ammonium chloride) is 53.49 g/mol.
The molar mass of NH3 (ammonia) is 17.03 g/mol.
The molar mass of H2O (water) is 18.02 g/mol.
The molar mass of CaCl2 (calcium chloride) is 110.98 g/mol.
First, we need to calculate the number of moles for each reactant:
Number of moles of CaO = mass of CaO / molar mass of CaO
= 112 g / 56.08 g/mol
= 2 moles
Number of moles of NH4Cl = mass of NH4Cl / molar mass of NH4Cl
= 224 g / 53.49 g/mol
= 4 moles
Next, based on the stoichiometric coefficients in the balanced equation, we can determine the number of moles for each product:
Number of moles of NH3 = 2 × number of moles of CaO (according to the balanced equation)
= 2 × 2 moles
= 4 moles
Number of moles of H2O = 2 × number of moles of CaO (according to the balanced equation)
= 2 × 2 moles
= 4 moles
Number of moles of CaCl2 = number of moles of NH4Cl (according to the balanced equation)
= 4 moles
Therefore, when the reaction is complete, there are:
- 2 moles of CaO
- 4 moles of NH4Cl
- 4 moles of NH3
- 4 moles of H2O
- 4 moles of CaCl2
To determine the number of moles of reactants and products in a chemical reaction, we need to follow these steps:
Step 1: Convert the given masses of the reactants (CaO and NH4Cl) into moles.
Step 2: Use stoichiometry to determine the mole ratios between the reactants and products.
Step 3: Convert the moles of products back into masses, if required.
Let's calculate the moles of each reactant and product:
1. Calculate the moles of CaO:
Moles of CaO = Mass of CaO / Molar mass of CaO
The molar mass of CaO is the sum of the atomic masses of calcium (Ca) and oxygen (O) found on the periodic table. The atomic masses are:
Ca: 40.08 g/mol
O: 16.00 g/mol
Molar mass of CaO = 40.08 + 16.00 = 56.08 g/mol
Moles of CaO = 112 g / 56.08 g/mol = 2.0 moles
2. Calculate the moles of NH4Cl:
Moles of NH4Cl = Mass of NH4Cl / Molar mass of NH4Cl
The molar mass of NH4Cl is the sum of the atomic masses of nitrogen (N), hydrogen (H), and chlorine (Cl) found on the periodic table. The atomic masses are:
N: 14.01 g/mol
H: 1.01 g/mol
Cl: 35.45 g/mol
Molar mass of NH4Cl = (14.01 + (4 * 1.01)) + 35.45 = 53.49 g/mol
Moles of NH4Cl = 224 g / 53.49 g/mol = 4.2 moles
3. Use the balanced equation to determine the mole ratios between the reactants and products:
From the equation: CaO + 2NH4Cl -> 2NH3 + H2O + CaCl2
The mole ratio between CaO and NH3 is 1:2, which means that for every 1 mole of CaO, 2 moles of NH3 are produced.
The mole ratio between CaO and H2O is 1:1, which means that for every 1 mole of CaO, 1 mole of H2O is produced.
The mole ratio between CaO and CaCl2 is also 1:1.
Based on these ratios, we can calculate the moles of products:
Moles of NH3 = 2 * Moles of CaO = 2 * 2.0 moles = 4.0 moles
Moles of H2O = 1 * Moles of CaO = 1 * 2.0 moles = 2.0 moles
Moles of CaCl2 = 1 * Moles of CaO = 1 * 2.0 moles = 2.0 moles
4. Finally, if required, convert the moles of products back into masses:
Mass of NH3 = Moles of NH3 * Molar mass of NH3
Mass of H2O = Moles of H2O * Molar mass of H2O
Mass of CaCl2 = Moles of CaCl2 * Molar mass of CaCl2
The molar mass of NH3 is the sum of the atomic masses of nitrogen (N) and hydrogen (H) found on the periodic table:
N: 14.01 g/mol
H: 1.01 g/mol
Molar mass of NH3 = 14.01 + (3 * 1.01) = 17.03 g/mol
Mass of NH3 = 4.0 moles * 17.03 g/mol = 68.12 g
The molar mass of H2O is the sum of the atomic masses of hydrogen (H) and oxygen (O):
H: 1.01 g/mol
O: 16.00 g/mol
Molar mass of H2O = (2 * 1.01) + 16.00 = 18.02 g/mol
Mass of H2O = 2.0 moles * 18.02 g/mol = 36.04 g
The molar mass of CaCl2 is the sum of the atomic masses of calcium (Ca) and chlorine (Cl):
Ca: 40.08 g/mol
Cl: 35.45 g/mol
Molar mass of CaCl2 = 40.08 + (2 * 35.45) = 110.98 g/mol
Mass of CaCl2 = 2.0 moles * 110.98 g/mol = 221.96 g
Therefore, when the reaction is complete, there are:
- 2.0 moles of CaO
- 4.2 moles of NH4Cl
- 4.0 moles of NH3
- 2.0 moles of H2O
- 2.0 moles of CaCl2