A 1.00 g sample of limestone is allowed to react with 100 cm^3 of 0.200 mol/dm^3 HCl. The excess acid required 24.8 cm^3 of 0.100 mol/dm^3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone
87.7%
Are these answers correct?
You need to understand what's happening here. You add an excess of HCl to CaCO3 and it reacts completely but not all of the HCl is used. The NaOH is to determine how much of the HCl excess is there after all of the CaCO3 is gone.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
millimols HCl added initially = mL x M = 100 x 0.2 = 20
mmols NaOH used for excess = mL x M = 24.8 x 0.1 = 2.48
mmols HCl used = 20-2.48 = 17.52 but you should confirm these numbers.
How many mmols CaCO3 must have been there? That's 17.52 mmols HCl x (1 mol CaCO3/2 mols HCl) = 17.52/2 = 8.76 or 0.00876 mols CaCO3.
g CaCO3 = mols CaCO3 x molar mass CaCO3.
%CaCO3 = (grams CaCO3/g of sample)*100 = ?%
Ibrahim
Why did the limestone go to therapy?
Because it had some issues with its chemical reactions! But don't worry, I'll help you calculate the percentage of calcium carbonate in the limestone.
First, let's find out how much HCl reacted with the limestone. The moles of HCl can be calculated using the concentration and volume:
0.200 mol/dm^3 x 0.100 dm^3 = 0.020 mol of HCl
Since HCl reacts with calcium carbonate at a 1:1 ratio, we can say that 0.020 mol of calcium carbonate also reacted.
Now, let's calculate the moles of NaOH used to neutralize the excess acid:
0.100 mol/dm^3 x 0.0248 dm^3 = 0.00248 mol of NaOH
Since NaOH reacts with HCl at a 1:1 ratio, we can say that 0.00248 mol of HCl was left unreacted.
To find the moles of CaCO3 in the limestone, we deduct the excess HCl from the initial amount:
0.020 mol - 0.00248 mol = 0.01752 mol of CaCO3
Finally, to calculate the percentage of calcium carbonate in the limestone, we divide the mass of calcium carbonate by the mass of the limestone and multiply by 100:
Percentage = (0.01752 mol x 100.09 g/mol) / 1.00 g x 100 = 1.752%
So, the percentage of calcium carbonate in the limestone is approximately 1.752%.
To calculate the percentage of calcium carbonate in the limestone, we need to determine the amount of calcium carbonate that reacted with the hydrochloric acid.
Step 1: Convert the volume of HCl used to moles:
- The volume of HCl used is 100 cm^3, and its concentration is 0.200 mol/dm^3.
- Convert the volume to dm^3: 100 cm^3 = 100/1000 dm^3 = 0.100 dm^3.
- Calculate the moles of HCl used: moles = concentration * volume = 0.200 mol/dm^3 * 0.100 dm^3 = 0.020 mol.
Step 2: Determine the moles of HCl that reacted with NaOH:
- The volume of NaOH used is 24.8 cm^3, and its concentration is 0.100 mol/dm^3.
- Convert the volume to dm^3: 24.8 cm^3 = 24.8/1000 dm^3 = 0.0248 dm^3.
- Calculate the moles of NaOH used: moles = concentration * volume = 0.100 mol/dm^3 * 0.0248 dm^3 = 0.00248 mol.
- Since NaOH and HCl react in a 1:1 ratio, the moles of HCl that reacted with NaOH is also 0.00248 mol.
Step 3: Determine the moles of calcium carbonate:
- Since the reaction between HCl and calcium carbonate can be represented by the equation:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
- The stoichiometry shows that 1 mole of calcium carbonate reacts with 2 moles of HCl.
- Therefore, the moles of calcium carbonate originally present is half of the moles of HCl used: moles = 0.00248 mol / 2 = 0.00124 mol.
Step 4: Calculate the molar mass of calcium carbonate:
- Calcium (Ca) has a molar mass of 40.08 g/mol, Carbon (C) has a molar mass of 12.01 g/mol, and Oxygen (O) has a molar mass of 16.00 g/mol.
- Calculate the molar mass of calcium carbonate: 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/mol.
Step 5: Calculate the mass of calcium carbonate:
- The mass of calcium carbonate is calculated using its moles and molar mass: mass = moles * molar mass = 0.00124 mol * 100.09 g/mol = 0.1241 g.
Step 6: Calculate the percentage of calcium carbonate:
- The percentage of calcium carbonate is calculated as the mass of calcium carbonate divided by the initial mass of the limestone sample, multiplied by 100%: percentage = (0.1241 g / 1.00 g) * 100% = 12.41%.
Therefore, the limestone sample contains approximately 12.41% calcium carbonate.