How many mL of 0.416 M of HCl are needed to dissolve 8.68 g of BaCO3?

2HCl(aq)+BaCO3(s)=BaCl2(aq)+H2O(l)+CO2(g)

BaCO3 + 2HCl ==> H2O + CO2 + BaCl2

mols CaCO3 = grams/molar mass = ?
mols HCl = 2x that. (Look at the coefficients in the balanced equation.)
M HCl = mols HCl/L HCl You have mols and M, solve for L HCl and convert to mL.

To determine the number of milliliters of 0.416 M HCl needed to dissolve 8.68 g of BaCO3, we first need to use stoichiometry to find the balanced equation for the reaction, and then calculate the moles of BaCO3 being reacted.

The balanced equation is:
2HCl(aq) + BaCO3(s) -> BaCl2(aq) + H2O(l) + CO2(g)

From the balanced equation, we can see that 2 moles of HCl react with 1 mole of BaCO3. This means that the molar ratio of HCl to BaCO3 is 2:1.

To find the moles of BaCO3, we can use the molar mass of BaCO3:

Molar mass of Ba = 137.33 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molecular mass of BaCO3 = (1 * 137.33) + (1 * 12.01) + (3 * 16.00) = 197.34 g/mol

Now we can calculate the moles of BaCO3:
moles of BaCO3 = mass of BaCO3 / molar mass of BaCO3
moles of BaCO3 = 8.68 g / 197.34 g/mol

Next, we need to determine the amount of HCl needed to react completely with the given moles of BaCO3. Since the molar ratio of HCl to BaCO3 is 2:1, we multiply the moles of BaCO3 by 2 to find the moles of HCl:

moles of HCl = 2 * moles of BaCO3

Finally, we can calculate the volume of 0.416 M HCl using the given concentration:

Molarity (M) = moles of solute / volume of solution in liters

Since we want the volume in milliliters (mL), we can rearrange the equation to:

Volume (mL) = (moles of solute / molarity) * 1000

Substituting the values, we have:

Volume of HCl = (moles of HCl / molarity of HCl) * 1000

Now you can plug in the values to calculate the volume of HCl required to dissolve 8.68 g of BaCO3.