Calcium carbonate reacts with hydrochloric acid according to the following reaction:
CaCO3 (s) + 2 HCl (aq) --> CaCl2 (aq) + H2O (l) + CO2 (g)
How many grams of calcium carbonate would be required to neutralize 50.0 cm3 of 8.00 M HCl?
To determine the number of grams of calcium carbonate required to neutralize a given volume and concentration of hydrochloric acid, you need to use the balanced chemical equation and perform a stoichiometric calculation. Here's how to solve the problem:
Step 1: Convert the given volume of hydrochloric acid to liters.
Since the volume was given in cm3, we need to convert it to liters. Remember that 1 cm3 is equal to 1 milliliter (mL), and 1 liter (L) is equal to 1000 milliliters. So, 50.0 cm3 is equal to 50.0 mL, which is equivalent to 0.0500 L.
Step 2: Calculate the number of moles of hydrochloric acid.
To find the number of moles, you can use the formula:
moles = volume (in liters) × concentration (in moles per liter)
Given that the concentration of HCl is 8.00 M and the volume is 0.0500 L:
moles of HCl = 0.0500 L × 8.00 mol/L
Step 3: Determine the stoichiometry between HCl and CaCO3.
From the balanced chemical equation, we know that the stoichiometric ratio between HCl and CaCO3 is 2:1. This means that 2 moles of HCl will react with 1 mole of CaCO3.
Step 4: Find the number of moles of CaCO3 required.
Since the stoichiometric ratio is 2:1, the number of moles of CaCO3 required will be half the number of moles of HCl:
moles of CaCO3 = (0.0500 L × 8.00 mol/L) / 2
Step 5: Calculate the mass of CaCO3.
To find the mass of CaCO3, you multiply the number of moles by the molar mass of CaCO3. The molar mass of CaCO3 is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms.
molar mass of CaCO3 = atomic mass of Ca + atomic mass of C + (3 × atomic mass of O)
Once you have the molar mass of CaCO3, you can calculate the mass:
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3
Step 6: Perform the calculations.
Now, you can substitute the values and solve to find the mass of CaCO3:
mass of CaCO3 = (0.0500 L × 8.00 mol/L) / 2 × molar mass of CaCO3
You can find the molar masses of the elements from the periodic table. The atomic mass of Ca is 40.08 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol.
Substituting the values and calculating will give you the mass of CaCO3 required to neutralize 50.0 cm3 of 8.00 M HCl.
Convert HCl to moles. moles = M x L.
Using the coefficients in the balanced equation, convert moles HCl to moles CaCO3.
Now convert mols CaCO3 to grams. g = moles x molar mass.