What mass (in grams) of calcium carbonate is needed to react completely with 73 mL of 0.10 M HCl according to 2HCl + CaCO3 --> CaCl2 + H20 + CO2?
This problem is worked the same way as the BaCl2/Na2SO4/BaSO4 problem. If you need extra help on stoichiometry, here is an example problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of calcium carbonate (CaCO3) needed to react with a given volume of hydrochloric acid (HCl), we can use the equation's balanced stoichiometry and molar concentrations.
The balanced equation shows that 2 moles of HCl react with 1 mole of CaCO3. Therefore, the mole ratio between HCl and CaCO3 is 2:1.
Step 1: Convert the given volume of HCl to moles:
Given volume of HCl = 73 mL
Concentration of HCl = 0.10 M
To convert mL to L, divide by 1000:
73 mL ÷ 1000 = 0.073 L
Now, using the molar concentration and volume, we can calculate the moles of HCl:
Moles of HCl = Concentration × Volume
Moles of HCl = 0.10 M × 0.073 L
Step 2: Determine the moles of CaCO3 needed using the mole ratio:
The mole ratio between HCl and CaCO3 is 2:1.
This means that for every 2 moles of HCl, we need 1 mole of CaCO3.
Moles of CaCO3 = Moles of HCl ÷ 2
Step 3: Calculate the mass of CaCO3 using its molar mass:
The molar mass of CaCO3 can be found by adding up the atomic masses of its components:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (there are 3 oxygen atoms in CaCO3)
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 × 16.00 g/mol)
Step 4: Calculate the mass of calcium carbonate:
Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3
By plugging in the calculated values, you can find the mass of calcium carbonate needed to react completely with the given volume of HCl.