A sample of impure Limestone weighing 0.3440 g was treated with 50 mL of HCl 0.1363 mol/l. after elimination of all CO2 by boiling,
excess acid was created by 4.45 mL of NaOH 0.1445 mol/L1. Calculate the% (w/w) of CaO and CaCO3 in the sample
To calculate the % (w/w) of CaO and CaCO3 in the sample, we'll first determine the moles of HCl and NaOH used in the reaction. Then, we'll use stoichiometry to find the moles of CaCO3 reacted and the moles of CaO formed. Finally, we'll calculate the % (w/w) of CaO and CaCO3.
Step 1: Calculate the moles of HCl used:
Moles of HCl = concentration (mol/L) x volume (L)
Moles of HCl = 0.1363 mol/L x 0.050 L = 0.006815 mol
Step 2: Calculate the moles of NaOH used:
Moles of NaOH = concentration (mol/L) x volume (L)
Moles of NaOH = 0.1445 mol/L x 0.00445 L = 0.000642725 mol
Step 3: Determine the moles of HCl reacted with CaCO3:
From the balanced chemical equation:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
we can see that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, the moles of HCl reacted with CaCO3 is half the moles of HCl used:
Moles of HCl reacted = 0.006815 mol / 2 = 0.0034075 mol
Step 4: Determine the moles of CaCO3 reacted and CaO formed:
From the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CaO:
Moles of CaCO3 reacted = Moles of CaO formed = 0.0034075 mol
Step 5: Calculate the molecular weight of CaO:
CaO = 40.08 g/mol
Step 6: Calculate the % (w/w) of CaO in the sample:
% (w/w) of CaO = (Moles of CaO formed x Molecular weight of CaO) / Sample weight
Sample weight = 0.3440 g
% (w/w) of CaO = (0.0034075 mol x 40.08 g/mol) / 0.3440 g = 0.3978 g/g x 100% = 39.78%
Step 7: Calculate the molecular weight of CaCO3:
CaCO3 = 100.09 g/mol
Step 8: Calculate the % (w/w) of CaCO3 in the sample:
% (w/w) of CaCO3 = (Moles of CaCO3 reacted x Molecular weight of CaCO3) / Sample weight
% (w/w) of CaCO3 = (0.0034075 mol x 100.09 g/mol) / 0.3440 g = 0.9884 g/g x 100% = 98.84%
Therefore, the % (w/w) of CaO in the sample is 39.78% and the % (w/w) of CaCO3 in the sample is 98.84%.
To calculate the % (w/w) of CaO (calcium oxide) and CaCO3 (calcium carbonate) in the sample, we need to follow a series of steps.
Step 1: Calculate the moles of HCl used in the reaction.
Given:
- Volume of HCl solution used: 50 mL
- Molarity of HCl solution: 0.1363 mol/L
To convert the volume of HCl solution used to moles of HCl, we use the following formula:
moles of HCl = volume (in L) × molarity
Converting the volume of HCl used to liters:
50 mL = 50/1000 = 0.05 L
Calculating the moles of HCl:
moles of HCl = 0.05 L × 0.1363 mol/L = 0.006815 mol
Step 2: Calculate the moles of NaOH used in the reaction.
Given:
- Volume of NaOH solution used: 4.45 mL
- Molarity of NaOH solution: 0.1445 mol/L
To convert the volume of NaOH solution used to moles of NaOH, we use the same formula as above:
moles of NaOH = volume (in L) × molarity
Converting the volume of NaOH used to liters:
4.45 mL = 4.45/1000 = 0.00445 L
Calculating the moles of NaOH:
moles of NaOH = 0.00445 L × 0.1445 mol/L = 0.000641525 mol
Step 3: Find the moles of HCl reacted with CaCO3.
The balanced chemical equation for the reaction between HCl and CaCO3 is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
From the equation, we can see that the molar ratio between CaCO3 and HCl is 1:2.
Since the moles of HCl reacted are twice the moles of CaCO3, we have:
moles of CaCO3 reacted = 0.006815 mol / 2 = 0.0034075 mol
Step 4: Calculate the moles of unreacted HCl.
Since excess NaOH was used, the remaining NaOH reacts with the unreacted HCl.
From the balanced equation, we know that the molar ratio between NaOH and HCl is 1:1.
Hence, the moles of unreacted HCl is equal to the moles of NaOH used:
moles of unreacted HCl = 0.000641525 mol
Step 5: Calculate the moles of CaO.
From the balanced chemical equation, we can see that:
1 mol of CaCO3 reacts to form 1 mol of CaO
moles of CaO = moles of CaCO3 reacted = 0.0034075 mol
Step 6: Calculate the mass of CaO and CaCO3.
We can use the molar mass of each compound to calculate their masses.
The molar mass of CaO (56.08 g/mol) and CaCO3 (100.09 g/mol) can be found in the periodic table.
mass of CaO = moles of CaO × molar mass of CaO = 0.0034075 mol × 56.08 g/mol = 0.1911 g
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3 = 0.0034075 mol × 100.09 g/mol = 0.3409 g
Step 7: Calculate the % (w/w) of CaO and CaCO3.
The % (w/w) of a compound is calculated using the following formula:
% (w/w) = (mass of compound / mass of sample) × 100
Given:
- Mass of impure limestone sample = 0.3440 g
Calculating the % (w/w) of CaO and CaCO3 in the sample:
% (w/w) of CaO = (0.1911 g / 0.3440 g) × 100 = 55.57%
% (w/w) of CaCO3 = (0.3409 g / 0.3440 g) × 100 = 99.06%
Therefore, the % (w/w) of CaO in the sample is approximately 55.57%, and the % (w/w) of CaCO3 in the sample is approximately 99.06%.