1. Ammonia gas can be prepared by the following unbalanced equation:
_____ CaO(s) + _____ NH4Cl (s) --> _____ NH3(g) + _____ H2O(g) + _____ CaCl2(s)
a. Balance the equation.
CaO + 2NH4Cl = 2NH3 + H2O + CaCl2
b. Suppose you begin this reaction with 112. grams of CaO and 224. grams of NH4Cl. Which is the limiting reagent?
c. What is the theoretical yield of NH3 in grams?
d. What mass of CaCl2 is also formed?
e. What mass of which reagent (starting material) is “left over” at the end of this process?
f. If the actual yield of NH3 is 55.6 grams, find the percent yield of NH3.
b. To determine the limiting reagent, we need to compare the amount of product that can be formed from each reactant.
First, we need to calculate the number of moles for each reactant:
The molar mass of CaO (calcium oxide) is 56.08 g/mol.
Number of moles of CaO = mass of CaO / molar mass of CaO
Number of moles of CaO = 112 g / 56.08 g/mol = 2 moles
The molar mass of NH4Cl (ammonium chloride) is 53.49 g/mol.
Number of moles of NH4Cl = mass of NH4Cl / molar mass of NH4Cl
Number of moles of NH4Cl = 224 g / 53.49 g/mol ≈ 4.19 moles
Now, we can compare the stoichiometric ratio of CaO to NH4Cl in the balanced equation. The coefficients in the balanced equation indicate that for every 1 mole of CaO, we need 2 moles of NH4Cl. Therefore, the actual ratio is:
Ratio of moles of CaO to NH4Cl = 2 moles / 1 mole ≈ 2
Since the moles of CaO (2 moles) are already greater than the moles of NH4Cl (4.19 moles), the limiting reagent is CaO.
c. To calculate the theoretical yield of NH3, we need to use the limiting reagent (CaO) to determine the maximum amount of product that can be formed.
From the balanced equation, the stoichiometric ratio of CaO to NH3 is 1:2.
Molar mass of NH3 (ammonia) is 17.03 g/mol.
Number of moles of NH3 = 2 moles of CaO × (1 mole of NH3 / 1 mole of CaO) = 2 moles
Theoretical yield of NH3 = number of moles of NH3 × molar mass of NH3
Theoretical yield of NH3 = 2 moles × 17.03 g/mol = 34.06 g
Therefore, the theoretical yield of NH3 is 34.06 grams.
d. The balanced equation tells us that 1 mole of CaO reacts to form 1 mole of CaCl2. Since the number of moles of CaO (2 moles) and CaCl2 (2 moles) are the same, the mass of CaCl2 formed would be equal to the mass of CaO:
Mass of CaCl2 = mass of CaO = 112 g
Therefore, the mass of CaCl2 formed is 112 grams.
e. To determine the mass of the reagent (starting material) that is left over, we need to calculate the number of moles consumed by the reaction. Since CaO is the limiting reagent, it will be completely consumed.
Number of moles of NH4Cl reacted = number of moles of CaO consumed = 2 moles
Mass of NH4Cl left over = remaining moles of NH4Cl × molar mass of NH4Cl
Mass of NH4Cl left over = (initial moles of NH4Cl - reacted moles of NH4Cl) × molar mass of NH4Cl
Mass of NH4Cl left over = (4.19 moles - 2 moles) × 53.49 g/mol ≈ 108.58 g
Therefore, the mass of NH4Cl left over is approximately 108.58 grams.
f. To find the percent yield of NH3, divide the actual yield by the theoretical yield and multiply by 100:
Percent yield of NH3 = (actual yield / theoretical yield) × 100
Percent yield of NH3 = (55.6 g / 34.06 g) × 100 ≈ 163.3%
Therefore, the percent yield of NH3 is approximately 163.3%.