what volume of 0.1 moldm-3 HCL acid would be required to dissolve 2.3 grams of calcium carbonate

equation CaCO3+2HCL----------=CaCl2+CO2+H20 (WATER)

I got 0.46 dm3 for my answer

To determine the volume of 0.1 moldm-3 HCl acid required to dissolve 2.3 grams of calcium carbonate (CaCO3), we need to use the equation provided:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

First, calculate the number of moles of calcium carbonate (CaCO3) using its molar mass (100.09 g/mol):

Moles of CaCO3 = mass / molar mass
Moles of CaCO3 = 2.3 g / 100.09 g/mol
Moles of CaCO3 = 0.023 mol

Now, based on the stoichiometry of the balanced equation, we know that 1 mol of CaCO3 reacts with 2 mol of HCl. Therefore, the number of moles of HCl required can be calculated as:

Moles of HCl = Moles of CaCO3 × (2 mol HCl / 1 mol CaCO3)
Moles of HCl = 0.023 mol × (2 mol HCl / 1 mol CaCO3)
Moles of HCl = 0.046 mol

Finally, we can calculate the volume of 0.1 moldm-3 HCl acid required using the equation:

Volume of HCl = Moles of HCl / concentration of HCl
Volume of HCl = 0.046 mol / 0.1 mol/dm^3
Volume of HCl = 0.46 dm^3

So, you would require 0.46 dm^3 (or 460 mL) of 0.1 moldm-3 HCl acid to dissolve 2.3 grams of calcium carbonate.

To find the volume of 0.1 moldm^-3 HCl acid required to dissolve 2.3 grams of calcium carbonate, you can use stoichiometry and the concept of molar concentration.

First, let's calculate the molar mass of calcium carbonate (CaCO3):
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol

Total molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol

Next, we need to calculate the number of moles of calcium carbonate:
moles = mass / molar mass
moles = 2.3 g / 100.09 g/mol ≈ 0.023 mol

From the balanced chemical equation, we can see that it takes 2 moles of HCl to react with 1 mole of CaCO3. Therefore, the number of moles of HCl needed to react with 0.023 mol of CaCO3 is:
moles of HCl = 2 * moles of CaCO3
moles of HCl = 2 * 0.023 mol = 0.046 mol

Now, we can calculate the volume of HCl using the molar concentration (0.1 moldm^-3):
volume = moles / molar concentration
volume = 0.046 mol / 0.1 moldm^-3 = 0.46 dm^3

Therefore, the volume of 0.1 moldm^-3 HCl acid required to dissolve 2.3 grams of calcium carbonate is 0.46 dm^3.

Work out the Mr for CaCO3, which is 100.1gmol-1. Use this to work out the number of moles using the equation n=m/Mr, which is 2.3g/100.1gmol-1 = 0.02977 mol.

Next work out the ration of CaCO3 to HCl which is 1:2. Therefore the number of moles of HCL is twice that of CaCO3 [0.02977 mol X 2 = 0.05954].
Then use the equation Volume = Moles / Concentration, which is 0.05954 mol / 0.1 moldm-3 = 0.5954 dm3, your answer; the volume of HCl.

Hope that's right!! :D