What volume of 3.00 mm hclhcl in liters is needed to react completely (with nothing left over) with 0.750 ll of 0.500 mm na2co3na2co3?

The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

The chemical equation for the reaction of sodium carbonate and HCl follows:

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

Hence, the volume of HCl needed is 0.250 L