Balanced equation: 2HCL+CaCO2>CaCl2+CO2+H2O

what volume of 6.00 M HCL is needed to completely react with .750 moles of CaCO3

To determine the volume of 6.00 M HCl needed to react with 0.750 moles of CaCO3, we need to use the balanced chemical equation and the stoichiometry between HCl and CaCO3.

From the balanced equation:
2 HCl + CaCO3 -> CaCl2 + CO2 + H2O

We can see that the stoichiometric ratio between HCl and CaCO3 is 2:1. This means that for every 2 moles of HCl, we need 1 mole of CaCO3.

Now, let's use the given information and convert from moles to volume.

First, calculate the number of moles of HCl needed to react with 0.750 moles of CaCO3:
Since the stoichiometric ratio is 2:1, we need half as many moles of HCl.

Moles of HCl = 0.750 moles CaCO3 / 2 = 0.375 moles HCl

Next, use the volume-molarity relationship to calculate the volume of 6.00 M HCl required.

Molarity (M) = moles / volume (L)

0.375 moles HCl = 6.00 M × volume (L)

Rearranging the equation, we get:

Volume (L) = 0.375 moles HCl / 6.00 M

Volume (L) ≈ 0.063 L or 63 mL (rounded to three significant figures)

Therefore, approximately 63 mL of 6.00 M HCl is needed to completely react with 0.750 moles of CaCO3.