The equation for the reaction between calcium carbonate and dilute nitric acid is shown.
Cacog(s) + 2HNO3(aq) + Ca(NO3)2(aq) + CO2(g) + H2O(1)
25g of calcium carbonate is reacted with an excess of dilute nitric acid.
Which mass of calcium nitrate and which volume of carbon dioxide is produced at room
temperature and pressure?

1- The first thing to do is to match the chemical reaction, so as to have the same amount of atoms of each element in both reagents and reaction products:

CaCO₃ (s) +2 HNO₃(aq) ⇒ Ca (NO₃) ₂ (aq) + CO₂ (g) + H₂O (l)

2- Then, the molar masses of the compounds that are given as data in the statement are calculated and the molar masses of the compounds that should be calculated are also calculated:

a) Calcium carbonate:

mCaCO₃ = mCa + mC + 3 x mO = 40.07g + 12.01g + 3 x 15.99g = 100.05 g / mol

b) Calcium nitrate:

mCa (NO₃) ₂ = mCa + 2 x (mN + 3 x mO) = 40.07g + 2x (14.00g + 3 x 15.99g) = 164.01 g / mol

c) Carbon dioxide:

mCO₂ = mC + 2 x mO = 12.01g + 2 x 15.99g = 43.99 g / mol

3- Now the calculations are carried out, according to the stoichiometric ratios of the reaction:

a) Calcium nitrate:

100.05 g CaCO₃ ____ 164.01 g Ca (NO₃)₂

25.0 g CaCO₃ _____ X = 40.98 g Ca (NO₃)₂

Calculation: 25.0 g x 164.01 g / 100.05 g = 40.98 g

40.98 g of calcium nitrate are produced

b) Carbon dioxide:

i- In this case, the number of moles that form the gas after the reaction must be calculated. The initial moles of Calcium Carbonate are calculated, and then proceed to make the stoichiometric calculation:

100.05 g CaCO₃ _____ 1 mol CaCO₃

25.0 g CaCO₃ _____ X = 0.25 mol CaCO₃

Calculation: 25.0 g x 1 mol / 100.05 g = 0.25 mol

So:

1 mol CaCO₃ _____ 1 mol CO₂

0.25 mol CaCO₃ ____ X = 0.25 mol CO₂

Calculation: 0.25 mol x 1 mol / 1 mol = 0.25 mol

ii- Now, considering that the ambient pressure is 760 torr (1 atm) and the ambient temperature is 25 ° C (298.15K), the general equation of ideal gases can be used to determine the volume of carbon dioxide obtained:

The equation being PxV = n x R x T

Where:

• P: Gas pressure in atm

• V: Volume of gas in L

• N: Number of moles of gas

• R: Gas constant, equal to 0.08206 L.atm / mol.K

• T: Temperature in K

Clearing in the equation the Volume, which is what you want to calculate, and replacing with the numerical values, you get:

V = (n x R x T) / P

V = (0.25mol x 0.08206 L.atm / mol.K x 298.15K) / 1atm = 6.12 L CO₂

Therefore, a volume of 6.12 L of carbon dioxide is obtained after the reaction occurs.