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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.