# How many milliliters of 0.306 M HCl are needed to react with 51.0 g of CaCO3?

##### 1 Answer

Jul 1, 2017

3330 mL

#### Explanation:

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: