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

1 Answer
Jul 1, 2017

Answer:

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#