What mass of calcium carbonate is needed to react with 20cm3 of 0.15M of nitric acid

I am so lost with this question. What do i need to find in order to get the mass of calcium carbonate.

CaCO3 + 2HNO3 = Ca(NO3)2 + H2O + CO2
Moles of HNO3 = 0.15*20/1000
= 0.003
Moles of CaCO3 = 0.003/2 = 0.0015
Mass = 0.0015 * 100
= 0.15 grams

CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2O(l) + CO2(g)

Find the moles of nitric acid that will react by multiplying the concentration by the volume (in liters):

(0.15 mol/L)(0.020 L) = 3.0x10^-3 mol

Using the balanced chemical equation above, we see that 1 mol of CaCO3 reacts with 2 mol of HNO3, so the moles of CaCO3 required is half that of HNO3:

(3.0x10^-3 mol HNO3)(1 mol CaCO3 / 2 mol HNO3) = 1.5x10^-3 mol CaCO3

You now have the required amount of moles of CaCO3 that will react with 20 mL of 0.15 M nitric acid. Now determine the required mass by multiplying the moles by the molar mass:

Molar mass of CaCO3 = 100.1 g/mol

Mass required = (1.5x10^-3 mol)(100.1 g/mol) = 0.150 g

∴ You require 0.15 g of calcium carbonate.