a. Write and balance the equation for the reaction of calcium carbonate with hydrochloric acid, then answer the following questions.

b. What mass of calcium carbonate is required to neutralize 83.5mL of 6.00M HCl?

does the balanced equation have any significance in this problem and for b. im not sure how to start it

CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

mols HCl = M HCl x L HCl = ?
Using the coefficients in the balanced equation, convert mols HCl to mols CaCO3. That will be mols HCl x (1 mol CaCO3/2 mols HCl) = mols HCl x 1/2 = ?
Then convert mols CaCO3 to grams. g CaCO3 = mols CaCO3 x molar mass CaCO3.

a. The balanced equation for the reaction of calcium carbonate (CaCO3) with hydrochloric acid (HCl) is as follows:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

This balanced equation is significant because it tells you the stoichiometry of the reaction, which means it gives you the ratio between the reactants and products. In this case, 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of calcium chloride, 1 mole of carbon dioxide, and 1 mole of water.

b. To determine the mass of calcium carbonate required to neutralize 83.5 mL of 6.00M HCl, you need to follow these steps:

1. Convert the given volume of hydrochloric acid to moles:

First, you need to find the number of moles of HCl present in the given volume using the equation:

moles = concentration (M) × volume (L)

Given: concentration (M) = 6.00 M and volume (L) = 83.5 mL = 0.0835 L

moles = 6.00 M × 0.0835 L = 0.501 moles

So, you have 0.501 moles of HCl.

2. Use the stoichiometry from the balanced equation to find the moles of calcium carbonate:

From the balanced equation, you know that 1 mole of calcium carbonate reacts with 2 moles of HCl. Therefore, the moles of calcium carbonate can be calculated as:

moles of CaCO3 = (moles of HCl) / 2

moles of CaCO3 = 0.501 moles / 2 = 0.2505 moles

So, you have 0.2505 moles of calcium carbonate.

3. Convert the moles of calcium carbonate to mass:

To find the mass of calcium carbonate, you need to use the molar mass of CaCO3. The molar mass of calcium carbonate is calculated by adding the atomic masses of calcium (Ca), carbon (C), and oxygen (O):

Molar mass of CaCO3 = (1 × atomic mass of Ca) + (1 × atomic mass of C) + (3 × atomic mass of O)

Molar mass of CaCO3 = (1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)

Molar mass of CaCO3 = 100.09 g/mol

Now, you can calculate the mass of calcium carbonate using the equation:

mass = moles × molar mass

mass = 0.2505 moles × 100.09 g/mol = 25.069 g

Therefore, the mass of calcium carbonate required to neutralize 83.5 mL of 6.00M HCl is 25.069 grams.