Chemistry Exercise

Question

How much mass% of CaCO3 does a piece of marble contain as at 25.0 °C and 1.013 atm, 900.0 mL of CO2 (gas) produced by the action of an excess HCl on 5.0 g of this marble?

thank u in advance

George Philips

Answer 1

Let's look at the balanced equation for what is taking place:

CaCO₃ + 2HCl ==> CaCl₂ + H₂O + CO₂

Now, using the Ideal Gas Law we can find out the number of moles (n) of CO₂ produced:

PV = nRT

n = PV/RT = (1.013 atm)(0.900 L)/(0.0821 Latm/Kmol)(298K)

n = 0.03726 moles CO₂

Since from the balanced equation 1 mol CO₂ = 1 mol CaCO₃, we know we had 0.03726 mol CaCO₃

Finding mass of CaCO₃ we obtain...

0.03726 mol CaCO₃ x 100 g CaCO₃/mol = 3.73 g CaCO₃

% mass = 3.73 g/5.0 g (x100%) = 74.6% CaCO₃ = 75%