How many milliliters of 1.237 M HCl are required to react with 27.02 grams of MgCO3 ?

Same procedure as above.

To determine the number of milliliters of 1.237 M HCl required to react with 27.02 grams of MgCO3, we need to use stoichiometry to find the moles of MgCO3 and then calculate the amount of HCl needed to react with those moles.

1. Convert the mass of MgCO3 (27.02 grams) to moles by using its molar mass. The molar mass of MgCO3 is calculated by summing the atomic masses of Mg (24.31 g/mol), C (12.01 g/mol), and 3O (3 x 16.00 g/mol).

Molar mass of MgCO3 = (24.31 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 84.31 g/mol

Moles of MgCO3 = Mass of MgCO3 / Molar mass of MgCO3
Moles of MgCO3 = 27.02 g / 84.31 g/mol

2. Next, we need to balance the chemical equation between HCl and MgCO3 to determine the stoichiometry. The balanced equation is:

MgCO3 + 2HCl -> MgCl2 + CO2 + H2O

From the balanced equation, we can see that 1 mole of MgCO3 reacts with 2 moles of HCl.

3. Use the moles of MgCO3 calculated in step 1 to determine the moles of HCl required. Since the stoichiometric ratio is 1:2 (MgCO3:HCl), the moles of HCl will be double the moles of MgCO3.

Moles of HCl = 2 x Moles of MgCO3

4. Convert the moles of HCl to milliliters by using the concentration of the HCl solution (1.237 M). The term "1.237 M" means that there are 1.237 moles of HCl in 1 liter of solution (1000 mL).

Moles of HCl = Volume (in liters) × Concentration (in moles/L)
Volume (in liters) = Moles of HCl / Concentration (in moles/L)

Since we want the volume in milliliters, we need to multiply the volume in liters by 1000.

Volume (in milliliters) = Volume (in liters) × 1000

Now, plug in the values from steps 1 to 4 to calculate the number of milliliters of 1.237 M HCl required to react with 27.02 grams of MgCO3.