a 110 g solid sample that is a mixture of CaCO3 and CaCl2 is reacted with 1.50 L of 1.45M HCl. Of the calcium salts, only calcium carbonate reacts with HCl- the reaction forms CaCl2 and CO2. The excess/unreacted HCl requires 0.850 L of 0.543 M NaOH to titrate it to the endpoint.
Determine the mass and mass percent of calcium carbonate in the original solid sample
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
1. mols HCl added = 1.5L x 1.45M = about 2 mol HCl but you need to do it more accurately. That reacts with all of the CaCO3 but doesn't use all of the HCl.
2. Excess HCl requires how many mols. That's 0.850L x 0.543M NaOH = about .46 mols NaOH.
3. mols HCl used = about 2 - 0.46 = about 1.5.
4. Use the coefficients in the balanced equation to convert to mols CaCO3. That is mols HCl x 1/2.
5. mass CaCO3 = mols x molar mass
6. %CaCO3 = (mass CaCO3/mass sample)*100 = ?
To determine the mass and mass percent of calcium carbonate in the original solid sample, we need to go through a series of steps. Let's break it down.
Step 1: Calculate the moles of HCl used in the reaction with calcium carbonate.
We have the volume and molarity of HCl, so we can use the formula:
moles = volume (L) x molarity (mol/L)
moles of HCl = 1.50 L x 1.45 mol/L
Step 2: Use the reaction stoichiometry to find the moles of calcium carbonate.
From the balanced equation, we know that 1 mole of calcium carbonate reacts with 2 moles of HCl. So, we need to multiply the moles of HCl by the stoichiometric ratio:
moles of CaCO3 = (moles of HCl) x (1 mole CaCO3 / 2 moles HCl)
Step 3: Calculate the mass of calcium carbonate.
We know the molar mass of calcium carbonate (CaCO3) is approximately 100.09 g/mol, so we can use the formula:
mass (g) = moles x molar mass
mass of CaCO3 = (moles of CaCO3) x (molar mass of CaCO3)
Step 4: Find the mass percent of calcium carbonate in the original sample.
The mass percent can be calculated using the formula:
mass percent = (mass of component / total mass of sample) x 100
mass percent of CaCO3 = (mass of CaCO3 / mass of sample) x 100
Now, let's do the calculations.
Given:
- Mass of sample = 110 g
- Volume of NaOH for titration = 0.850 L
- Molarity of NaOH = 0.543 mol/L
Step 1:
moles of HCl = 1.50 L x 1.45 mol/L = 2.175 mol HCl
Step 2:
moles of CaCO3 = (2.175 mol HCl) x (1 mol CaCO3 / 2 mol HCl) = 1.088 mol CaCO3
Step 3:
mass of CaCO3 = (1.088 mol CaCO3) x (100.09 g/mol CaCO3) = 108.92 g CaCO3
Step 4:
mass percent of CaCO3 = (108.92 g CaCO3 / 110 g sample) x 100 = 98.928%
Therefore, the mass of calcium carbonate in the original solid sample is approximately 108.92 g, and the mass percent of calcium carbonate in the sample is approximately 98.928%.