Question

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

Answer 1

3330 mL

Explanation:

From the chemical equation, we can see that for every mole of `CaCO_3`, 2 moles of HCl react. This means that if you want them to react exactly in proportion, the number of moles of HCl needed is double that of `CaCO_3`.

`n(HCl)=2*n(CaCO_3)`

The molar mass (MM) of calcium carbonate is:

`MM(CaCO_3)=40.1+12.0+3*16.0=100.1" "g/(mol)`

The number of moles of calcium carbonate can now be calculated:

`n(CaCO_3)=(mass)/(MM)=51.0/100.1=0.509" "mol`

So, now we double the answer to get the amount of HCl needed:

`n(HCl)=2*n(CaCO_3)=2*0.509=1.02" " mol`

Rearranging the concentration formula we can get the volume needed in litres:

`C=n/VrArrV=n/C`

`V=1.02/0.306=3.33" "L`

Convert to mL:

`V=3.33*10^3" "mL`