25 g of Iron (II) Oxide reacts with 25 g of Chloric acid to produce Iron (II) chlorate and water.
a. Determine the limiting reactant
b. Determine the excess reactant
c. Determine the theoretical yield of water
d. Determine how much excess reactant is left
e. Determine the percent yield of water, if the actual value of water is 2.4 g.
0.9866 g
To determine the limiting reactant, we need to compare the number of moles of each reactant and calculate the number of moles of product that each can produce.
a. Determine the limiting reactant:
1. Calculate the molar mass of Iron (II) Oxide (FeO):
- Iron (Fe) molar mass = 55.845 g/mol
- Oxygen (O) molar mass = 15.999 g/mol
- FeO molar mass = 55.845 + 15.999 = 71.844 g/mol
2. Calculate the number of moles of Iron (II) Oxide:
- Moles = Mass / Molar mass
- Moles of FeO = 25 g / 71.844 g/mol = 0.348 mol
3. Calculate the molar mass of Chloric acid (HClO3):
- Hydrogen (H) molar mass = 1.008 g/mol
- Chlorine (Cl) molar mass = 35.453 g/mol
- Oxygen (O) molar mass = 15.999 g/mol
- HClO3 molar mass = 1.008 + 35.453 + (3 * 15.999) = 84.461 g/mol
4. Calculate the number of moles of Chloric acid:
- Moles of HClO3 = 25 g / 84.461 g/mol = 0.296 mol
5. To determine the limiting reactant, compare the moles of each reactant. The reactant that produces fewer moles of product is the limiting reactant.
- The moles of Iron (II) chlorate (Fe(ClO3)2) produced would be equal to the moles of Chloric acid used, as per the balanced equation.
Balanced equation:
3HClO3 + FeO -> Fe(ClO3)2 + H2O
Ratio of moles:
3 : 1 :: 3 : 1
Since the moles of Chloric acid (0.296 mol) and moles of Iron (II) Oxide (0.348 mol) are very close, we can see that the limiting reactant is Chloric acid (HClO3).
b. Determine the excess reactant:
In this case, Iron (II) Oxide (FeO) is the excess reactant.
c. Determine the theoretical yield of water:
1 mol of Chloric acid (HClO3) produces 1 mol of water (H2O) as per the balanced equation. Therefore, the theoretical yield of water is equal to the moles of Chloric acid used.
Theoretical yield of water = 0.296 mol
d. Determine how much excess reactant is left:
To calculate the excess reactant left, we need to determine the moles of the limiting reactant used.
Moles of FeO used = Moles of HClO3 used = 0.296 mol
Moles of FeO left = Moles of FeO initially - Moles of FeO used
Moles of FeO left = 0.348 mol - 0.296 mol = 0.052 mol
e. Determine the percent yield of water:
The percent yield of water can be calculated using the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (2.4 g / (0.296 mol x molar mass of water)) x 100
Molar mass of water (H2O) = 1.008 + 1.008 + 15.999 = 18.015 g/mol
Percent yield = (2.4 g / (0.296 mol x 18.015 g/mol)) x 100
To determine the limiting reactant, excess reactant, theoretical yield of water, and the amount of excess reactant left, we need to follow a series of steps. Let's go through each step:
Step 1: Write the balanced chemical equation:
FeO + 2HCl → FeCl2 + H2O
Step 2: Calculate the molar masses:
FeO: 55.85 g/mol + 16 g/mol = 71.85 g/mol
HCl: 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Step 3: Convert the mass of each reactant to moles:
moles of FeO = 25 g / 71.85 g/mol ≈ 0.348 mol
moles of HCl = 25 g / 36.46 g/mol ≈ 0.686 mol
Step 4: Compare the moles of reactants to the coefficients in the balanced equation.
From the balanced equation, the molar ratio between FeO and HCl is 1:2. This means that 1 mole of FeO reacts with 2 moles of HCl.
Since the mole ratio of FeO to HCl is 1:2, we can deduce that FeO is the limiting reactant. This is because we have 0.348 mol of FeO, which is less than the 2 × 0.348 = 0.686 mol of HCl.
Therefore, FeO is the limiting reactant and HCl is the excess reactant.
Step 5: Determine the theoretical yield of water:
From the balanced equation, we see that the stoichiometry between FeO and water is 1:1. This means that 1 mol of FeO reacts to produce 1 mol of water.
Since we have 0.348 mol of FeO, the theoretical yield of water can be calculated as follows:
theoretical yield of water = 0.348 mol × (1 mol H2O / 1 mol FeO) × (18 g/mol H2O) ≈ 6.26 g
Therefore, the theoretical yield of water is approximately 6.26 g.
Step 6: Determine the amount of excess reactant left:
To do this, we need to calculate the moles of excess HCl that reacted with the limiting amount of FeO:
moles of excess HCl reacted = moles of HCl - (moles of FeO × (2 mol HCl / 1 mol FeO))
moles of excess HCl reacted = 0.686 mol - (0.348 mol × (2 mol HCl / 1 mol FeO)) ≈ 0.535 mol
Step 7: Calculate the mass of excess HCl left:
mass of excess HCl left = moles of excess HCl left × molar mass of HCl
mass of excess HCl left = 0.535 mol × 36.46 g/mol ≈ 19.5 g
Therefore, approximately 19.5 g of HCl is left as the excess reactant.
Step 8: Calculate the percent yield of water:
Percent yield can be calculated using the formula:
percent yield = (actual yield / theoretical yield) × 100
Given that the actual value of water is 2.4 g and the theoretical yield is 6.26 g, we can calculate the percent yield as follows:
percent yield = (2.4 g / 6.26 g) × 100 ≈ 38.3%
Therefore, the percent yield of water is approximately 38.3%.