How many milliliters of 0.306 M HCl are needed to react with 51.0 g of CaCO3?
From the chemical equation, we can see that for every mole of
The molar mass (MM) of calcium carbonate is:
The number of moles of calcium carbonate can now be calculated:
So, now we double the answer to get the amount of HCl needed:
Rearranging the concentration formula we can get the volume needed in litres:
Convert to mL: